hydraulics topic 2 uniform flow in open channel
TRANSCRIPT
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Hydraulics
Topic 2. Uniform Flow in Open Channel
Dr. Mohd Ariff bin Ahmad [email protected]
Assoc. Prof. Dr. Tan Lai [email protected]
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Learning Outcomes
At the end of this topic, students should be able to:
i. Understand the concept of uniform flow
ii. Calculate normal flow depth in variable channelsections using Chezy and Manning equations
iii. Determine the best hydraulic/effective section of
open channel
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Open Channel Flow
Classification
based on Time
Classification
based on Space
Steady Unsteady Uniform Non-Uniform
GVF RVF
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• Uniform flow is considered to be steady only, sinceunsteady uniform flow is practically does not exist.
• Steady uniform flow is rare in natural streams, only
happens in prismatic channels.
• We adopt / assume uniform flow for most flow
computations because uniform flow calculation is
simple, practical and provide satisfactory solution.
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The 132 km long All-American Canal links California's Imperial Valley to the Colorado River. This new
concrete-lined section saves about 3.8 million of water a year over its leaky earthen forerunner
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The concrete channel of Los Angeles River (NGM, 2010)
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The Klang River, Kuala Lumpur & Selangor
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• In uniform flow, the normal depth y o occurs when depth of water is
the same along the channel.
• Normal depth y o implies that the water depth, flow area, wetted
perimeter, velocity and discharge at every section of the channel
are constant within a prismatic channel.
• Thus, in uniform flow, the energy line, water surface and channel
bottom are parallel, i.e. the slopes are equal S f = Sw = So = S.
g
V
2
2
y o
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2.1 Velocity Distribution
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Natural channel
0.84
0.820.800.76
0.70
0.620.48
V max
V max
0.52
0.50
0.450.400.35
y o
0.53
Rectangular channel
V average
0.6y o
V max0.2y
o
V
y
y o
Velocity distribution
Depends on the geometry of the channel and wetted boundary
roughness
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2.2 Chezy and Manning Equations
Two most common equations used in the uniform flow computations:
1. Chezy formula
2. Manning formula
2
1
2
1
oSCRV
2
1
3
21
oSRnV
x
o
x SRV constantThus, the general uniform flow equation:
C = Chezy roughness coefficient
n = Manning roughness coefficient
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6
11R
nC Difference between Chezy and Manning formulae
Factors determining the roughness are surface roughness, vegetation,
channel irregularity, channel alignment, silting and scouring,
obstruction, size and shape of channel, stage and discharge, seasonalchange, and suspended material and bed load.
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The Chezy two assumptions are:
1. The force resisting the flow per unit area of the channel bed is
proportional to the square of the velocity:
2. The effective component of the gravity force causing the flow
must be equal to the total force of resistance. This is also the
basic principle of uniform flow where uniform flow will be
developed if the resistance is balance by the gravity forces:
Derivation of Chezy equation
PLV k F f 2
sin ALF g
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g
V
2
2
y o
W
Datum
A
P
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sin2 ALPLV k
o
ALSPLV k 2
oSP
A
k V
2
2
1
2
12
1
oSRk V
2
1
2
1
oSRC V where C = Chezy coefficient
1221 sin MM pF W p f
Since for uniform flow, 2121 and MM p p
Total force of resistance is counter-balances with the
effective component of gravity, which is acts parallel
to the channel bed.
Fr = Force of resistance
W = Weight of the fluid = ALθ = Slope angle of the bed
= Specific weight of the fluid
A = Cross sectional area of the channel
L = Characteristic length of the channel
The resistance to flow is proportional to
the square of the velocity.
Fr = resistance to flow (N)
Aw = wetted area = PxL
P = wetted perimeter
L = length of the channel
K = constant of proportionality
V = mean velocity of flow
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A rectangular channel 2.0 m wide carries water at a depth of 0.5 m.
The channel is laid on a slope of 0.0004. The Chezy coefficient is 73.6.Compute the discharge of the channel.
Given B = 2.0 m, y = 0.5 m, So = 0.0004 and C = 73.6
A = By = 1 m2, P = B + 2y = 3 m, R = 1/3 m
B
y
oRS AC Q
0004.0
3
16.731 Q
/sm850.0 3Q
Activity 2.1
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Water flows in a triangular channel with side slope 1.5(H) : 1(V),
bottom slope 0.0002 and Chezy coefficient of 67.4. The depth offlow is 2.0 m. Find the flow rate and average velocity. Based on
Froude number, determine the state of flow.
z
y1
Given y = 2.0 m, z = 1.5, So = 0.0002 and C = 67.4
A = zy 2 = 6 m2, P = 2y = 7.211 m, R = A/P = 0.832 m, D = A/T = 6/2zy = 1 m
Activity 2.2
oRSC V
0002.0832.04.67 V
m/s869.0V
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AV Q
869.06Q
/sm217.5 3Q
gDV Fr
181.9
869.0Fr
flowlsubcritica277.0Fr
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Chezy resistance factor C
The following two equations can be used to determine Chezycoefficient:
1. Ganguillet-Kutter
2. Bazin
Rn
S
nSC
o
o
00155.0231
100155.023
R
mC
1
87
n = Kutter coefficient
m = Bazin coefficient
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Table 2.1a Values of Manning roughness coefficient n
Surface characteristics Range of n
(a) Lined channels with straight alignment
Concrete
i. formed, no finish 0.013 - 0.017
ii. trowel finish 0.011 - 0.015
iii. float finish 0.013 - 0.015
iv. gunite, good section 0.016 - 0.019
v. gunite, wavy section 0.018 - 0.022
Concrete bottom, float finish, sides as indicated
i. dressed stone in mortar 0.015 - 0.017
ii. random stone in mortar 0.017 - 0.020
iii. cement rubble masonry 0.020 - 0.025iv. cement rubble masonry, plastered 0.016 - 0.020
v. dry rubble (rip-rap) 0.020 - 0.030
Tile 0.016 - 0.018
Brick 0.014 - 0.017
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Table 2.1b Values of Manning roughness coefficient n
Surface characteristics Range of n
Sewers (concrete, asbestos-cement, vitrified-clay
pipes)
0.012 - 0.015
Asphalt
i. smooth 0.013
ii. rough 0.016
Concrete lined, excavated rock
i. good section 0.017 - 0.020
ii. irregular section 0.022 - 0.027
Laboratory flumes-smooth metal bed, glass or
perspex sides
0.009 - 0.010
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Manning roughness coefficient n
= 0.020 - 0.022
Manning roughness coefficient n
= 0.020 - 0.022
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Manning roughness coefficient n = 0.022 - 0.024
Manning roughness coefficient n
= 0.020
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Surface characteristics Range of n
(b) Unlined, non-erodible channels
Earth, straight and uniform
i. clean, recently completed 0.016 - 0.020
ii. clean, after weathering 0.018 - 0.025
iii. gravel, uniform section, clean 0.022 - 0.030
iv. with short grass, few weeds 0.022 - 0.033
Channels with weeds and brush, uncut
i. dense weeds, high as flow depth 0.050 - 0.120
ii. clean bottom, brush on sides 0.040 - 0.080
iii. dense weeds or aquatic plants in deep
channels
0.030 - 0.035
iv. grass, some weeds 0.025 - 0.033
Rock 0.025 - 0.045
Table 2.1c Values of Manning roughness coefficient n
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Surface characteristics Range of n
(c) Natural channels
Smooth natural earth channels, free from growth,
little curvature
0.020
Earth channels, considerably covered with small
growth
0.035
Mountain streams in clean loose cobbles, rivers
with variable section with some vegetation on thebanks
0.040 - 0.050
Rivers with fairly straight alignment, obstructed
by small trees, very little under brush
0.060 - 0.075
Rivers with irregular alignment and cross-section,
covered with growth of virgin timber and
occasional patches of bushes and small trees
0.125
Table 2.1d Values of Manning roughness coefficient n
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Manning roughness coefficient n = 0.11
Manning roughness coefficient n = 0.20
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Grassed swale
Table 2.2 Values of Manning
roughness coefficient for
grassed swale
Surface
cover
Manning n
Short grass 0.030 - 0.035
Tall grass 0.035 - 0.050
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Table 2.3 Proposed values of Bazin coefficient m
Description of channel Bazin coefficient m
Very smooth cement of planed wood 0.11
Unplaned wood, concrete, or brick 0.21
Ashlar, rubble masonry, or poor brickwork 0.83
Earth channels in perfect condition 1.54
Earth channels in ordinary condition 2.36
Earth channels in rough condition 3.17
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Calculate the velocity and discharge in a trapezoidal channel having a
bottom width of 20 m, side slopes 1(H) : 2(V), and a depth of water 6
m. Given Kutter's n = 0.015 and So = 0.005.
z
y1
B
Activity 2.3
Given B = 20 m, y = 6.0 m, z = 0.5, So = 0.005 and n = 0.015
A = By + zy 2 = 138 m2,
P = B + 2y = 33.42 m,
R = A/P = 4.13 m
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R
n
S
nSC
o
o
00155.0
231
100155.023
Ganguillet-Kutter
13.4
015.0
005.0
00155.0231
015.01
005.000155.023
C
769.76C
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oRSC V Chezy velocity
005.013.4769.76 V
m/s03.11V
AV Q Discharge
03.11138Q
/sm14.1522 3Q
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Find the equivalent Bazin coefficient m for the question in Activity 2.3
and compare the Chezy coefficients obtained from Kutter n & Bazin m.
Assume that for concrete with Kutter n = 0.015, Bazin m = 0.21
R
mC 1
87
Bazin
Known A = 138 m2, P = 33.42 m, R = 4.13 m
13.4
21.01
87
C
Kutter)-Ganguillet(from76.769Bazin)(from852.78 C
Activity 2.4
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A trapezoidal channel is 10.0 m wide and has a side slope of
1.5(H) : 1(V). The bed slope is 0.0003. The channel is lined with
smooth concrete n = 0.012. Compute the mean velocity and
discharge for a depth of flow of 3.0 m.
z
y1
B
Given B = 10 m, y = 3.0 m, z = 1.5, So = 0.0003 and n = 0.012
A = By + zy 2 = 43.5 m2,
P = B + 2y = 20.817 m,
R = A/P = 2.090 m
21 z
Activity 2.5
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2
1
3
2
1oSR
nV Manning velocity
2
1
3
2
0003.0090.2012.0
1V
m/s359.2V
AV Q Discharge
359.25.43
/sm625.102 3
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In the channel of Example 2.5, find the bottom slope necessary to
carry only 50 m3/s of the discharge at a depth of 3.0 m.
Activity 2.6
Given B = 10 m, y = 3.0 m, z = 1.5 and n = 0.012
and A = 43.5 m2, P = 20.817 m, R = 2.090 m
2
1
3
21
oS ARnQ Manning discharge
2
1
3
2
09.25.43012.0
150 oS
0000712.0oS
51012.7 oS
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A triangular channel with an apex angle of 75 carries a flow of
1.2 m3/s at a depth of 0.80 m. If the bed slope is 0.009, find the
roughness coefficient C and n of the channel.
Activity 2.7
Given y = 0.80 m, So = 0.009, = 75, and Q = 1.2 m3/s
and A = zy 2 = 0.491 m2, P = 2y = 2.017 m,
R = A/P = 0.2435 m
21 z
z
y1 75
2
tan
z
2
75tan
0.767
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2
1
3
21
oS AR
n
Q Using Manning equation
2
1
3
2
009.02435.0491.01
2.1 n
0151.0n
2
1
2
1
oSCARQ Using Chezy equation
2
1
2
1
009.02435.0491.02.1 C
197.52C
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A trapezoidal channel of bottom width 25 m and side slope
2.5(H):1(V) carries a discharge of 450 m3/s with a normal depth of
3.5 m. The elevations at the beginning and end of the channel are
685 m and 650 m, respectively. Determine the length of the
channel if n = 0.02.
Given B = 25 m, z = 2.5, y o = 3.5, n = 0.02, and Q = 450 m3/s
z
y1
B
A = By + zy 2 = 118.125 m2
P = B + 2y = 43.848 m 21 z
R = A/P = 2.694 m
Activity 2.8
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2
1
3
21
oS AR
n
Q Manning equation,
2
1
3
2
694.2125.11802.0
1450 oS
00155.0oS
H
oL
zS
HL65068500155.0
m13.22601HL
Manning equation,
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2.3 Conveyance
Conveyance K of a channel section is a measure of the carrying
capacity of the channel section per unit longitudinal slope. It is
directly proportional to discharge Q.
1. Chezy formula
2. Manning formula
2
1
2
1
oSCARQ
2
1
3
21
oS ARnQ
2
1
CARK
3
21 ARnK
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Section factor Z in the Manning formula is AR2/3, which is a function of
the depth of flow.
In Manning formula 21
3
21
oS ARnQ
Therefore,2
132
oS
Qn AR
Section factor AR2/3 is normally used to compute the normal depth y o
when the discharge Q, bottom slope So and Manning roughnesscoefficient n are provided.
Computation of y o could be through either direct trial-and-error
computation, based on graph, or through provided design chart.
2.4 Section Factor
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Activity 2.9
A trapezoidal channel 5.0 m wide and having a side slope of 1.5(H) :
1(V) is laid on a slope of 0.00035. The roughness coefficient n = 0.015.Find the normal depth for a discharge of 20 m3/s through this channel.
Given B = 5.0 m, z = 1.5, So = 0.00035, n = 0.015, and Q = 20 m3/s
z
y1
B
A = By + zy 2 = 5y + 1.5y 2
P = B + 2y = 5 + 2 y 21 z 25.3
y
y y
P
AR
25.325
5.15 2
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Arranging Manning equation as a function of section factor,
2
13
2
oS
Qn AR
2
1
3
21
oS ARnQ Manning equation,
2
1
3
2
22
00035.0
015.020
25.325
5.155.15
o
oooo
y
y y y y
036.16
25.325
5.15
32
3
52
o
oo
y
y y
Therefore, y o = 1.820 m
y o (m)
32
3
52
25.325
5.15
o
oo
y
y y
By trial-and-error:
1
2
1.8
1.820
5.391
19.159
15.706
16.035
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Graphically,
036.16
25.325
5.15
3
2
3
52
o
oo
y
y y
y o (m)
32
3
52
25.325
5.15
o
oo
y
y y
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 )
y o = 1.82 m
16.036
Therefore, y o = 1.820 m
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Design Chart is available,
Circular
Rectangular ( z = 0)
B
3
8
3
2
3
8
3
2
and
od
AR
B
AR
o
d
y
B
y and
0.2194
0.37
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036.1632
AR
3
8
3
8
32
5
036.16
B
AR2194.0
37.0B
y
537.0 y
Therefore, y o = 1.85 m
At the x -axis,
Intersecting at z = 1.5 of trapezoidal channel gives
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Design chart for lined open drain fromUrban Stormwater Management Manual
for Malaysia (Department of Irrigation
and Drainage, 2000)
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Activity 2.10
A concrete-lined trapezoidal channel with n = 0.015 is to have a
side slope of 1(H) : 1(V). The bottom slope is to be 0.0004. Find thebottom width of the channel necessary to carry 100 m3/s of
discharge at a normal depth of 2.50 m.
z
y1
B
Given y o = 2.5 m, z = 1, So = 0.0004, n = 0.015, and Q = 100 m3/s
A = By + zy 2 = 2.5B + 6.25
P = B + 2y = B + 7.071 21 z
071.7
25.65.2
B
B
P
A
R
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Manning equation as a function of section factor,
2
13
2
oS
Qn AR
2
1
3
2
0004.0
015.0100071.725.65.225.65.2
BBB
75
071.7
25.65.2
3
2
3
5
B
B
By trial-and-error, B = 16.33 m
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Activity 2.11
Water flows uniformly at 10 m3/s in a rectangular channel with a base
width of 6.0 m, channel slope of 0.0001 and Manning's coefficient n =0.013. Using trial-and-error method, find the normal depth.
B
y
Given Q = 10 m3/s, B = 6.0 m, So = 0.0001 and n = 0.013
A = By = 6y
P = B + 2y = 6 + 2y
y
y R
3
3
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2
1
3
2
oS
Qn
AR
2
1
3
2
0001.0
013.010
3
36
o
ooy
y y
167.23
3 32
o
ooy
y y
By trial-and-error, y o = 1.942 m
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A sewer pipe of 2.0 m diameter is laid on a slope of 0.0004 with
n = 0.014. Find the depth of flow when the discharge is 2 m3
/s.
2 r
D
y o
sin228
2
D
Area A =
Perimeter P = D
Activity 2.12
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2
13
2
oS
Qn AR Manning equation:
2
13
2
0004.0
014.02 AR
3
8
3
8
3
2
2
4.1
D
AR
2205.0
3
8
3
2
D
AR
For design chart:
6.0D
y o
26.0 oy = 1.20 m
Intersecting at circular section gives
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Design Chart:
Circular
Rectangular (z = 0)
B
3
8
3
2
3
8
3
2
and
od
AR
B
AR
od
y
B
y and
0.2205
0.6
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Simplification for Wide Rectangular Channel
Wide channel: 02.0B
y o
For wide channel, is small, thereforeB
y ooy R
Or simply, oy R
Discharge per unit width
Normally used in rectangular channels.
B
Qq Discharge per unit width
Unit is m3/s/m.
yV q or
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Water flows through a very wide channel at a rate of 2.5 m3/s/m.
The channel has a base width of 60 m, channel slope of 0.005 and
Manning's coefficient of 0.013. What is the normal depth?
Given: q = 2.5 m3/s/m, B = 60 m, So = 0.005, n = 0.013
For a wide rectangular channel, R = y
V y q oManning equation:
Activity 2.13
2
1
3
21
oo SRny q
2
1
3
51
oo Sy nq
2
1
3
5
005.0013.0
15.2 oy
m6272.0oy
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2.5 Best Hydraulic Section (Most Effective Section)
3
21
ARnK
A non-erodible channel should be designed for the best hydraulic
efficiency.
Best hydraulic section gives minimum area for a given discharge.
Referring to the channel conveyance,
for a constant flow area A, the conveyance increases with increase
in hydraulic radius R or decrease in the wetted perimeter P.
Simply, Qmax, Rmax and P min gives best hydraulic section.
Pmin - reduces construction cost (less lining material), and
- reduces friction force.
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Cross
section
Side
slope zArea A
Wetted
perimeter P
Hydraulic
radius R
Top width
T
Hydraulic
depth D
Section
factor Z
Trapezoid
Rectangle - 2y
2
4y 2y y
Triangle 1 y 2 2y
Semicircle - 2y
Parabola -
Table. Best hydraulic sections
23y
2
2y
2
3
24y
y 32
y 22
y
y 3
28
2
y
2
y
4
2y
2
y
2
y
y 3
34
y 22
y 4
3
2
y
y 4
y 3
2
5.2
2
3y
5.22y
5.2
2
2y
5.2
4y
5.2
9
38y
3
1
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What is the best hydraulic section for a rectangular channel?
B
y
By AFor a rectangular channel,
y BP 2
Let's first assume A to be constant:
y y AP 2
22 y
A
dy
dP
For best hydraulic section 0d
d
y
P
Activity 2.14
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For best hydraulic section 022
ey
A
22 ey A
2
2 ee y By ey B 2
ey BP 2ee y y P 22
ey P 4
P
A
R
e
e
y
y R
4
2 2
2
ey
R
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Show that the best hydraulic trapezoidal section is one-half
of a hexagon.
60 1
3
1z
For a trapezoid,
2zy By A
212 zy BP
Activity 2.15
Let's first assume A and z to be constant:
For best hydraulic section 0d
d
y
P
zy y
AB
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22 12 zzy A
dy dP
212 zy zy
y
AP Substituting B
For best hydraulic section 012 2
2 zz
y
A
e 22
12 ey zz A
212 zy zy
y
AP And,
zzy P e 2122
P
AR Therefore,
zzy zzy R
e
e
2
22
12212
2
ey R
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If z is allowed to vary,2212 ey zz A
zz
A
y e 212
zzy P e 2122Substitute into P,
zzzz A
P
2
2 12122
zz AP 2122
0d
d
z
P
3
1ez
When
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zzy P e 2122
2
12 zy BP
212 zy PB
ey B3
2
ey P 32
2212 ey zz A
23 ey A
3
1ezWhen ,
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A slightly rough brick-lined trapezoidal channel carrying a
discharge of 25.0 m3
/s is to have a longitudinal slope of 0.0004.Analyse the proportions of
(a) an efficient trapezoidal channel section having a side of
1.5(H) : 1(V),
(b) the most efficient-channel section of trapezoidal shape.
Rough brick-lined gives Manning roughness n = 0.017
Activity 2.16
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(a) Fixed side slopes of 1.5(H) : 1(V),
For best hydraulic section2212 ey zz A
2
ey R
21056.2 ey A
From Manning equation, 21
3
21
oS AR
n
Q
2
13
2
20004.0
21056.2
017.0
125
eey
y
m8298.2ey 2
zy By A
e
e
e y y
y B 5.1
1056.2 2
m7137.1B 1.52.830 m1
1.714 m
and
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(b) If the side slope is not fixed, the side slope and other channel
characteristics for most-efficient trapezoidal section are
31ez
ey B3
2
23 ey A 2
ey R
5774.0ez
m045.3ey
m516.3eB
From Manning equation, 21
321oS AR
nQ
2
13
2
20004.0
21056.2
017.0
125
eey
y
0.5774
3.045 m1
3.516 m
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2.6 Channels of Compound Sections
Compound sections channel - channels that are composed of several
distinct subsections with each subsection different in roughness fromothers.
Manning equation is applied separately to each subsection to
determine the mean velocity.
n
i
i i AV Q1
A
SK
V
o
n
i
i 2
1
1
Or
A i i 2 17
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[Final exam question, Semester I, Session 2013/2014]
A composite channel as shown is designed to convey 19.8 m3/s of
water. The channel on a longitudinal slope So = 1:2000 is to belined with concrete (n = 0.017). Determine the normal depth of
flow based on graphical method.
4 m
3
23 m
Activity 2.17
A i #2
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Assignment #2
Q1. [Final Exam Sem II, Session 2008/2009]
(a) What is conveyance factor K ?
(b) Figure Q1(b) shows a compound channel and its dimensions.
The channel has bottom slope of 0.0036 and side slope of 1.5(H)
: 0.75(V). Determine the value of Chezy resistance coefficient C
and velocity of flow if flowrate is 10 m3
/s.
1.5 m
0.2 m
0.5 m
Figure Q1(b)
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Q1. (c) A very wide rectangular channel has a slope of 0.0004 and
Manning n = 0.02. If 2.54 m3/s/m flow is to be conveyed in this
channel, estimate the normal depth.
(d) A trapezoidal channel is to carry 18 m3/s of flowrate on a
bottom slope of 0.0009. Given that Manning's n is 0.026 and the
sides of channel are inclined 63.44° to the vertical, determine
the bottom width, depth and velocity for the best hydraulic
section.
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Q2. [Final Exam Sem I, Session 2010/2011]
(a) Utilizing the concept of section factor, prove that the section in
Figure Q2(a) gives
when the discharge of the uniform flow is 33.6 m3/s, bed slope
So = 0.001 and Manning coefficient n = 0.015.
94.1512.8
1058.41058.4
3
22
2
y
y y y y Z
y o
y o
2y o
45
60 10 m
Figure Q2(a)
2
oy
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Q2. (b) Determine the depth of flow y o of the channel if the best
hydraulic section is needed for a composite section as in Figure
Q2(b) to convey 6.5 m3/s of flow. Manning coefficient n and bed
slope are 0.015 and 0.0015, respectively.
4.5 m
y o
y 1
y 2
Figure Q2(b)
Q3 [Final Exam Sem I Session 2007/2008]
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Q3. [Final Exam, Sem I, Session 2007/2008]
(a) Water flows at a depth of 2.5 m in a rectangular concrete
channel (n = 0.013) of width 12 m and bed slope 0.0028. Find
the velocity and rate of flow.
(b) A housing area needs a channel to convey 9.8 m3/s of runoff. A
trapezoidal channel is proposed with 3 m width and side slope
3(horizontal) : 4(vertical). If the channel is concrete-lined (n =
0.013) and bottom slope So is 1 : 2000, determine the normal
depth using graphical method.
Q4. [Final Exam, Sem I, Session 2007/2008]
(a) Prove that the most efficient cross section for triangular channel
is half of a square.
(b) A concrete-lined irrigation channel with Manning's n = 0.020 isneeded to convey 12.5 m3/s of flow. The channel has a
trapezoidal section with bottom slope So = 0.0015. Determine
the most effective size of the channel if the side slope is
restricted to 3(horizontal) : 1(vertical).
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