7/29/2019 Chapter 9 Uniform Flow in Open Channel
1/33an Afnizan b Wan Mohamed
KAP/FKAAS
By:-
MR WAN AFNIZAN BIN WAN MOHAMED
DEPT. OF WATER & ENVIRONMENTAL ENGINEERING
FAC. OF CIVIL & ENVIRONMENTAL ENGINEERING
e-mail: [email protected]
HYDRAULICSHYDRAULICSHYDRAULICSHYDRAULICS
(DFC 2053)(DFC 2053)(DFC 2053)(DFC 2053)
CHAPTER 9CHAPTER 9CHAPTER 9CHAPTER 9
UNIFORM FLOW IN
OPEN CHANNEL
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CONTENT
CHANNEL GEOMETRY
TYPES OF FLOW
DISCHARGE FORMULA
APPLICATION OF NORMAL
DEPTH : CHART APPROACH
BEST EFFECTIVE SECTION
STATE OF FLOW
TYPES OF FLOW
Two criteria :-
A TIME AS THE CRITERION
(i) Steady Flow
Depth of flow does not change with time
(ii) Unsteady Flow
Depth of flow changes with time
B SPACE AS THE CRITERION
(i) Uniform Flow
Depth of flow is same at every section of channel
(ii) Non Uniform Flow
Depth of flow changes along of channel section
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.... Cont.... Cont.... Cont.... Cont
TYPES OF FLOW
B SPACE AS THE CRITERION
(iii) Rapidly Varied Flow
Depth of flow changes abruptly over a comparativelyshort distance (e.g : Hydraulic jump)
(iv) Gradually Varied Flow
Depth of flow changes gradually over a comparativelylong distance
C COMBINATION OF TWO CRITERION
(i) Steady Uniform
(ii) Steady Rapidly Varied
(iii) Steady Gradually Varied
.... Cont.... Cont.... Cont.... Cont
TYPES OF FLOW
C COMBINATION OF TWO CRITERION
(iv) Unsteady Rapidly Varied
(v) Unsteady Gradually Varied
Open Channel Flow
Steady Flow Unsteady Flow
Uniform Flow Non Uniform Flow
Gradually Varied FlowRapidly Varied Flow
Figure 9.1 : Types of flow in open channel
Note : Unsteady Uniform
Flow Not exist
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STATE OF FLOW
Influenced by three factors :-
Inertial Force
Viscosity Force
Gravity Force
A EFFECT OF INERTIAL & VISCOSITY FORCES
)(RNumbereynoldsRForceViscosity
ForceInertiale====
or represent as :-
====
==== vRvRRe
. 9.1
STATE OF FLOW
State of flow ( based on Reynolds Number ) :-
.... Cont.... Cont.... Cont.... Cont
where ;
v = Velocity (m/s)
= Dynamic viscosity (N.s/m2) = Water density (1000 kg/m3)= Kinematic viscosity (m2/s)R = Hydraulic radius (m)
P
AR ====
. 9.2
(i) Re 500 Laminar Flow
Flow : Slow & Shallow
Slippery surface
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STATE OF FLOW
Cont .. State of flow ( based on Reynolds Number ) :-
.... Cont.... Cont.... Cont.... Cont
(ii) Re 12500 Turbulent Flow
Flow : Normally occurred in open channel
Transitional Flow(iii) 500 Re 12500
STATE OF FLOW
B EFFECT OF INERTIAL & GRAVITY FORCES
)(FNumberFroudeForceGravity
ForceInertialr====
or represent as :-
. 9.3
.... Cont.... Cont.... Cont.... Cont
gD
vFr ====
where ;
v = Velocity (m/s)
g = Earths gravity (m/s2)
D = Hydraulic depth (m)
T
AD ==== . 9.4
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STATE OF FLOW
State of flow ( based on Froude Number ) :-
.... Cont.... Cont.... Cont.... Cont
(i) Fr = 1 Critical Flow
(ii) Fr < 1 Sub Critical Flow
Super Critical Flow(iii) Fr >1
Flow in slow condition
Flow in turbulent condition
CHANNEL GEOMETRY
Three types of channel :-
A NATURAL CHANNEL
Figure 9.2 : Natural channel
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.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
B ARTIFICIAL CHANNEL
Figure 9.3 : Artificial channel
.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
C PRISMATIC CHANNEL
Uniform cross section & slope at whole channel
length.
Usually artificial channel.
CHANNEL GEOMETRY ELEMENT
y = Depth of water (m)T = Top width water surface (m)B = Base width water surface (m)P = Wetted perimeter (m)A = Wetted area (m2)R = Hydraulic radius R = A/P (m)D = Hydraulic depth D = A/T (m)
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.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
T
B
Py
Figure 9.4 : Channel geometric element
.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
Figure 9.5 : Channels sides slope
I SIDES SLOPE, Z
1
z
Note : If slope, = 45 z = 1( z at left & right side is same )
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.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
Figure 9.5 : Channels sides slope
II CHANNEL SLOPE, So ( unit less )
.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
III RELATIONSHIP BETWEEN v & Q
AVQ ==== . 9.5where ;
Q = Discharge or flow rate (m3/s)
v = Velocity (m/s)
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.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENTDERIVATION OF CHANNEL FORMULA
B
y
1
z
1
z
31
2
L
Figure 9.6 : Derivation of channel formula
.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
From Figure 9.6 :
T = Top width water surface
zy2BT ++++====
A = Wetted area
A = Area 1 + Area 2 + Area 3
(((( )))) (((( )))) (((( )))) (((( ))))[[[[ ]]]] (((( )))) (((( ))))
2zyByA
yzy2
1yByzy
2
1A
++++====
++++++++
====
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.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
Cont . From Figure 9.6 :
P = Wetted perimeter
(((( ))))
2
22
222
22
z1yL
z1yL
yzyL
zyyL
++++====
++++====
++++====
++++====
.... Cont.... Cont.... Cont.... Cont
CHANNEL GEOMETRY
CHANNEL GEOMETRIC ELEMENT
Cont . From Figure 9.6 :
P = Wetted perimeter
2
z1y2BP
L2BP
++++++++====
++++====
therefore ;
Note :Use this trapezoidal formula (A, T & P) to find formulae forrectangular & triangular shape.
For Rectangular z = 0 for Triangular B = 0
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.... Cont.... Cont.... Cont.... Cont CHANNEL GEOMETRY
B + 2zyBy + zy2
2zyzy2
B + 2yBBy
PTASHAPE
B
y
T
zz11 y
T
y1z
T
B
1z
d
T
y
2z1y2 ++++
2z1y2B ++++++++
)sin(8
d2
2
sind
2
d
( in radian ) ( in radian )( in angle )
To sum up .. Table 9.1 : Channels geometric elements
LetLetLetLets take as take as take as take a
break!!!break!!!break!!!break!!!
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Based on the figure given find :-
i) Top width water surface (T), wettedarea (A), wetted perimeter (P) &hydraulic radius (R).
ii) If Q = 2.4 m3/s, determine the flowstate.
iii) If inclined length (L) = 50 m, find thecost to construct this channel (Given
excavation cost = RM 3/m3
and liningcost = RM 5/m2)
EXAMPLE 9.1EXAMPLE 9.1EXAMPLE 9.1EXAMPLE 9.1
3 m
2 m
1 m
60
.... Cont.... Cont.... Cont.... Cont
SOLUTION:SOLUTION:SOLUTION:SOLUTION:
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SOLUTIONSOLUTIONSOLUTIONSOLUTION
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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Water flows 0.8 m depth inside a 1.2 mdiameter culvert. Calculate top widthwater surface (T), wetted area (A),wetted perimeter (P) and hydraulic radius(R).
EXAMPLE 9.2EXAMPLE 9.2EXAMPLE 9.2EXAMPLE 9.2
SOLUTION:SOLUTION:SOLUTION:SOLUTION:
SOLUTIONSOLUTIONSOLUTIONSOLUTION
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
DISCHARGE FORMULA
Several discharge formula includes :
oRSACQ ==== Chezy formula
oSARn
1Q 3
2
==== Manning formula
Where ;
A = Channel area (m2)
C = Chezys coefficient (m0.5/s)
n = Mannings coefficient
R = Hydraulic radius (m)
So = Channel slope
Most popular !!
. 9.6
. 9.7
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.... Cont.... Cont.... Cont.... Cont DISCHARGE FORMULA
0.013xii) Unreap Wood
0.012xi) Reap Wood
0.012x) Steel
0.016ix) Aspalt (rough)
0.013viii) Aspalt (smooth)
0.015vii) Masonry (inside cement mortar)
0.013vi) Stone (covered)
0.017v) Concrete
0.013iv) Cement (mortar)
0.011iii) Cement (plane / smooth)
0.027 0.035ii) Earth ground (vegetation)
0.022i) Earth ground (clean)
Artificial Channel:
0.040 0.050iii) Mountain River
0.100ii) Vegetation
0.030i) Clean and Straight
Natural Channel:
MANNING VALUE,nTYPES OF CHANNEL SURFACE
Water flows inside trapezoidal channelhaving sides slope 1.5, base width 2.5 m,channel slope 0.0016 and Manningscoefficient 0.012. If normal depth insidethe channel is 3 m, what is the discharge?
EXAMPLE 9.3EXAMPLE 9.3EXAMPLE 9.3EXAMPLE 9.3
SOLUTION:SOLUTION:SOLUTION:SOLUTION:
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SOLUTIONSOLUTIONSOLUTIONSOLUTION
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
APPLICATION OF NORMAL DEPTH
Normal depth determination approaches :
Trial & Error
Graph
ChartOnly this part will be
focused in this course !!
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APPLICATION OF NORMAL DEPTH
.... Cont.... Cont.... Cont.... Cont
CHART METHOD
Section factor, Zo is divided with B8/3 ( B = Channel width )
or d8/3 ( D = Channel diameter )
Limited if channel is triangular
Formula used :-
3
8
o3
8
3
2
BS
nQ
B
AR====
3
8
oo3
8
o
32
dS
nQ
d
AR====
Rectangular &
Trapezoidal shape
Circular shape
. 9.8
. 9.9
APPLICATION OF NORMAL DEPTH.... Cont.... Cont.... Cont.... Cont
Figure 9.7 : Determination of normal depth using chart method
38oBS
nQ====
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Water flows through rectangular channelat 10 m3/s having base width 6 m, channelslope 0.0001 and Mannings coefficient0.013. Find normal depth using chartmethod ?
EXAMPLE 9.4EXAMPLE 9.4EXAMPLE 9.4EXAMPLE 9.4
SOLUTION:SOLUTION:SOLUTION:SOLUTION:
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
BEST EFFECTIVE SECTION
Shape that will be studied :
Rectangular
Trapezoidal
Triangular
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What
is
Effective Section
BEST EFFECTIVE SECTION
.... Cont.... Cont.... Cont.... Cont
BEST EFFECTIVE SECTION Can convey Dischargemaximum (Qmax) or Hydraulic Radius maximum (Rmax) withWetted Perimeter minimum (Pmin)
How it is derived ??? P is differentiated from Manningformula.
BEST EFFECTIVE SECTION
oSARn
1Q 3
2
====
o
3
2S
P
A
n
1Q
35
====
Note :A, n & So value - fix
Attempt to get Qmax Pmin 0y
PminP ====
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.... Cont.... Cont.... Cont.... Cont
BEST EFFECTIVE SECTION
Advantages of P minimum :
Reduced flow friction
Reduced construction cost
Get P equation in
y value
How to determine best effective section ???
y
P
0
y
P====
Trapezoidal
Triangular
Rectangular
TOP WIDTH
(T)
HYDRAULIC
RADIUS
(R)
WETTED
PERIMETER
(P)
AREA
(A)
SHAPE
.... Cont.... Cont.... Cont.... Cont BEST EFFECTIVE SECTION
To sum up . (this formula need to remember !!)
2y2 y42
yy2
2
y2y2
4
2y
3y2
y2
3y22
y
3
3y4
3/1z ==== For another z or sides s lope angle use this formulato get new A valuezy2z1y2B
2++++====@ = 60
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Water flows 14.5 m3/s inside rectangularchannel having channel slope 0.002 andMannings coefficient 0.017. Find the besteffective section for this channel.
EXAMPLE 9.5EXAMPLE 9.5EXAMPLE 9.5EXAMPLE 9.5
SOLUTION:SOLUTION:SOLUTION:SOLUTION:
SOLUTIONSOLUTIONSOLUTIONSOLUTION
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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Water flows 3.45 m3/s inside trapezoidalchannel having sides slopes angle 45,channel slope 1:1200 and Manningscoefficient 0.014. Find the best effectivesection for this channel.
EXAMPLE 9.6EXAMPLE 9.6EXAMPLE 9.6EXAMPLE 9.6
SOLUTION:SOLUTION:SOLUTION:SOLUTION:
SOLUTIONSOLUTIONSOLUTIONSOLUTION
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SOLUTIONSOLUTIONSOLUTIONSOLUTION
SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont
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