Download Channel Design Design of Channels

Channel Design
Hydromechanics VVR090
Design of Channels
• lined channels – minimizing lining material costs
• unlined channels – maximum permissible velocity and
threshold of movement (stable hydraulic section)
Concrete-lined
channel
Unlined channel
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Objectives of Channel Design
Transport water between two points in a safe and cost-effective
manner. Includes economical, safety, and esthetics aspects.
Here, mainly hydraulic aspects are considered.
General observations:
• Conveyance of a channel increases with the hydraulic radius
(wetted perimeter deceases). From Manning’s formula.
• The best hydraulic section is a semicircle (for a given area it
has the minimum wetted perimeter).
• For a specific cross section, the proportion that produce the
best hydraulic section (maximum flow) might be derived.
• The best hydraulic section might not be the best from an
economical point of view.
Best Hydraulic Cross Section I
Maximize the flow for a given cross-sectional shape and area.
From the Manning formula:
Q=
1
AR 2 / 3 So1/ 2
n
Qn
= AR 2 / 3
So
Thus:
⎛ Qn
⎞
max ⎜
= AR 2 / 3 ⎟ should be found for a given A
⎜ S
⎟
⎝ o
⎠
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Best Hydraulic Cross Section II
Optimum geometries for different cross sections.
Rectangular cross section: best hydraulic cross
section when the water depth is half the channel
width.
Practical Considerations for Best Hydraulic Section
• The area needed to excavate the best hydraulic section might
be larger than the area required to achieve the flow area
• It may not be possible to construct a stable best hydraulic
section in the natural material
• The cost of excavation depends on other things than the
amount of material removed (e.g., access to the site, cost of
disposing material)
• The slope of the channel must also be considered.
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Channel Design – General Aspects
1. The minimum permissible velocity is the lowest velocity that
will prevent sedimentation and vegetative growth (crude
estimates: 0.6 – 0.9 m/s for sedimentation and 0.75 m/s for
vegetation).
2. The side slopes depend primarily on the engineering
properties of the material through which the channel is
excavated.
3. The freeboard refers to the vertical distance between either
the top of the channel or the top of the channel lining and the
water surface (design flow at normal depth).
Side Slopes
4
Channel Freeboard I
Lined channel
Channel Freeboard II
Preliminary estimate of freeboard (unlined channel):
F = Cy
F: freeboard (ft)
y: design depth (ft)
C: coefficient (=1.5 for 20 ft3/s and 2.5 for 3000 ft3/s)
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Flow Around a Channel Curve I
Flow Around a Channel Curve II
Rise in water surface at the outer bank:
Δh =
u 2b
gR
b: channel width
R: distance from center of curve to centerline of channel
Take into account velocity variation:
Δh = 2.3
⎛R ⎞
u2
log ⎜ o ⎟
g
⎝ Ri ⎠
Ri: inner radius
Ro: outer radius
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Use of Lined Channels
Lined channels may be used:
• to permit transmission of water at high velocities through
areas of deep or difficult excavation
• to permit transmission of water at a reduced construction
cost
• to decrease channel seepage
• to reduce operation and maintenance costs
• to ensure stability of the channel section
Typical Lined Channel Cross Sections
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Design Procedure for Lined Channels
Minimize the cost of the lining material. Identical to find best
hydraulic section, if uniform thickness of the lining material is
employed.
Design procedure:
1. Estimate n
2. Compute the value of the section factor
3. For appropriate expression for A and R, compute yN
4. Compute channel properties
5. Check minimum permissible velocity and Froude number
6. Estimate height of lining above surface and freeboard
7. Summarize results with dimensioned sketch
Design for Different Lining Thickness or Material
The base of the channel and the sides of the channel might be
lined with different material of same material with different
thickness.
Design procedure involves minimizing the total cost (C) of the
lining material:
min ( C ) = min ( Cb + Cs )
Cb: material cost for channel base (per unit length)
Cs: material cost of sides (per unit length)
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Design of Stable, Unlined, Earthen Channels
Find a stable cross section =>
One where neither scour nor deposition constitutes a problem.
Three types of unstable sections:
1. Scouring occur but no deposition
2. Deposition occur but no scouring
3. Both scour and deposition occur
Maximum Permissible Velocities
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Theoretical Approach to Stable Cross Section I
Bed shear stress (balance between gravity and force due
to flow resistance):
τo = ρgRSo
For a wide channel, R ≈ y N , giving:
τo = ρgy N So
The shear stress is not uniformly distributed along the
perimeter. Complex problem to determine correct distribution,
but for trapezoidal cross section the following applies:
τomax = ρgy N So
τ
max
o
= 0.76ρgy N So
maximum along bottom
maximum along side slopes
Maximum Unit Tractive Force
Channel side
Channel bottom
(in terms of ρgy N So )
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Theoretical Approach to Stable Cross Section II
Study a particle in equilibrium, just before mobilization.
Employ a balance between the shear force mobilizing the
sediment and the resisting force depending on the material.
Balance for particles along the bed:
τ L Ae = Ws tan α
τL =
Ws
tan α
Ae
Ae: effective area
Ws: submerged particle weight
a: angle of friction (= angle of repose)
Angle of Respose
The angle to the horizontal at which
grains start to roll on a flat bed of
sediment that is gradually tilted
from the horizontal.
A representative value on the angle of
repose is 32 deg.
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Theoretical Approach to Stable Cross Section III
Balance for particles on the side slopes:
( τS Ae ) + (Ws sin Γ )
2
2
= Ws cos Γ tan α
Ws
tan 2 Γ
cos Γ tan α 1 −
τS =
tan 2 α
Ae
Tractive force ratio:
K=
τS
tan 2 Γ
sin 2 Γ
= cos Γ 1 −
=
1
−
τL
tan 2 α
sin 2 α
Angle of Repose
tana
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Maximum Unit Tractive Force for
Different Materials
non-cohesive
cohesive
Typical Channel Cross Sections
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Design Procedure for Unlined Channels
Find a stable cross section. Design procedure:
1. Estimate n
2. Estimate angle of repose for channel material
3. ...
Consult French for the procedure.
Example 14.265: Water Surface Profile
Upstream an Obstruction
Water flowing at the normal depth in a rectangular concrete
channel that is 12 m wide encounters an obstruction (see figure),
causing the water level to rise above the normal depth at the
obstruction and for some distance upstream. The water discharge
is 126 m3/s and the channel bottom slope is 0.00086. The depth of
water just upstream from the obstruction is 4.55 m. Find the
distance upstream to the point where the surface is at the normal
depth.
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Normal water depth:
Q=
1
AR 2 / 3 S 1/ 2
n
⎛ 12 y N ⎞
1
126 =
12 y N ⎜
⎟
0.013
⎝ 12 + 2 y N ⎠
⎛ 12 y N ⎞
2.256 ⋅ ⎜
⎟
⎝ 12 + 2 y N ⎠
2/3
−
2/3
0.000861/ 2
10.5
=0
yN
y N = 2.95 m (by trial and error)
Critical water depth:
1/ 3
⎛ q2 ⎞
ycr = ⎜ ⎟
⎝ g ⎠
1/ 3
⎛ 1 ⎛ 126 ⎞ 2 ⎞
=⎜
⎜ 9.81 ⎜⎝ 12 ⎟⎠ ⎟⎟
⎝
⎠
= 2.24 m
yN > ycr => subcritical flow at normal water depth
Start at obstruction and calculate upstream through the step
method.
Δxi
(y+u
=
2
/ 2g )
i +1
2 2
− ( y + u2 / 2g )
So − ( n u / R
4/3
)
i
i +1/ 2
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Table for Step Calculation
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