Pertemuan 17 - 18
Open Channel 1
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Open Channel Flow
Uniform Open Channel Flow is the hydraulic condition in which the
water depth and the channel cross section do not change over some
reach of the channel
Through experimental observations and calculations, Manning’s
Equation was developed to relate flow and channel geometry to
water depth. Knowing the flow in a channel, you can solve for the
water depth. Knowing the maximum allowable depth, you can solve
for the maximum flow.
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Open Channel Flow
Manning’s equation is only accurate for cases where the cross sections of a
stream or channel are uniform. Manning’s equation works accurately for man
made channels, but for natural streams and rivers, it can only be used as an
approximation.
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Manning’s Equation
Terms to know in the Manning’s equation:
V = Channel Velocity
A = Cross sectional area of the channel
P = Wetted perimeter of the channel
R = Hydraulic Radius = A/P
S = Slope of the channel bottom (ft/ft or m/m)
n = Manning’s roughness coefficient
n = 0.015 for concrete n = 0.03 for clean natural
channel n = .01 for glass
Yn = Normal depth (depth of uniform flow)
Area
Yn
Y
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Wetted Perimeter
X
Slope = S = Y/X
Manning’s Equation
V = (1/n)R2/3√(S)
for the metric system
V = (1.49/n)R2/3√(S)
for the English system
Q = A(k/n)R2/3√(S)
k is either 1 or 1.49
As you can see, Yn is not directly a part of Manning’s equation. However,
A and R depend on Yn. Therefore, the first step to solving any
Manning’s equation problem, is to solve for the geometry’s cross
sectional area and wetted perimeter:
For a rectangular Channel
Yn
B
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Area = A = B x Yn
Wetted Perimeter = P = B + 2Yn
Hydraulic Radius = A/P = R = BYn/(B+2Yn)
Simple Manning’s Example
A rectangular open concrete (n=0.015) channel is to be designed to
carry a flow of 2.28 m3/s. The slope is 0.006 m/m and the bottom
width of the channel is 2 meters.
Determine the normal depth that will occur in this channel.
First, find A, P and R
A = 2Yn
P = 2 + 2Yn
R = 2Yn/(2 + 2Yn)
Next, apply Manning’s equation
Q = A(1/n)R2/3√(S) 
2.28 = (2Yn)x(1/0.015)x(2Yn/(2 + 2Yn))2/3x√(0.006)
Solving for Yn
Yn
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2m
Yn = 0.47 meters
The Trapezoidal Channel
House flooding occurs along Brays Bayou when water
overtops the banks. What flow is allowable in Brays Bayou
if it has the geometry shown below?
Slope
S = 0.001 ft/ft
25’
Concrete Lined
n = 0.015
35’
Θ = 20°
A, P and R for Trapezoidal Channels
A = Yn(B + Yn cot θ)
P = B + (2Yn/sin θ )
Yn
θ
B
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R = (Yn(B + Yn cot θ)) / (B + (2Yn/sin θ ))
The Trapezoidal Channel
Slope
S = 0.0003 ft/ft
25’
Concrete Lined
n = 0.015
35’
Θ = 20°
A = Yn(B + Yn cot θ)
A = 25( 35 + 25 x cot(20)) = 2592 ft2
P = B + (2Yn/sin θ )
P = 35 + (2 x 25/sin(20)) = 181.2 ft
R = (Yn(B + Yn cot θ)) / (B + (2Yn/sin θ ))
R = 2592’ / 181.2’ = 14.3 ft
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The Trapezoidal Channel
Slope
S = 0.0003 ft/ft
25’
Concrete Lined
n = 0.015
35’
Θ = 20°
A = 2592 ft2 R = 14.3 ft
Q = A(1.49/n)R2/3√(S)
Q = 2592 x (1.49 / .015) x 14.32/3 x √(.0003)
Q = Max allowable Flow = 26,273 cfs
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Manning’s Over Different Terrains
S = .005 ft/ft
5’
5’
5’
3’
Grass
n=.03
Concrete
n=.015
Grass
n=.03
3’
Estimate the flow rate for the above channel?
Hint:
Treat each different portion of the channel separately. You
must find an A, R, P and Q for each section of the channel
that has a different roughness coefficient.
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Manning’s Over Different Terrains
S = .005 ft/ft
5’
5’
5’
3’
Grass
n=.03
Concrete
n=.015
Grass
n=.03
3’
The Grassy portions:
A = 5’ x 3’ = 15 ft2
For each section:
P = 5’ + 3’ = 8 ft R = 15 ft2/8 ft = 1.88 ft
Q = 15(1.49/.03)1.882/3√(.005)
Q = 80.24 cfs per section  For both sections…
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Q = 2 x 80.24 = 160.48 cfs
Manning’s Over Different Terrains
S = .005 ft/ft
5’
5’
5’
3’
Grass
n=.03
A = 5’ x 6’ = 30 ft2
Concrete
n=.015
Grass
n=.03
3’
The Concrete portions
P = 5’ + 3’ + 3’= 11 ft
R = 30 ft2/11 ft = 2.72 ft
Q = 30(1.49/.015)2.722/3√(.005)
Q = 410.6 cfs
For the entire channel…
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Q = 410.6 + 129.3 = 540 cfs
Uniform Open Channel Flow
Basic relationships
Continuity equation
Energy equation
Momentum equation
Resistance equations
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Flow in Streams
Introduction
 Effective Discharge
 Shear Stresses
 Pattern & Profile

 Open Channel Hydraulics
 Resistance Equations
 Compound Channel
• Sediment Transport
• Bed Load Movement
• Land Use and Land Use Change
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Continuity Equation
Inflow
3
3a
A
Change in Storage
3b
Outflow
1
A
2
Section AA
Inflow – Outflow = Change in Storage
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General Flow Equation
Q = va
Equation 7.1
Area of the
cross-section
Flow rate
(cfs) or (m3/s)
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Avg. velocity
of flow at a
cross-section
(ft/s) or (m/s)
(ft2) or (m2)
Resistance (velocity) Equations
Manning’s Equation
Equation 7.2
Darcy-Weisbach Equation
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Equation 7.6
Velocity Distribution In A Channel
Depth-averaged velocity is above
the bed at about 0.4 times the depth
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Manning’s Equation
• In 1889 Irish Engineer, Robert Manning presented the formula:
1.49 2 3 1 2
v
R S
n
Equation 7.2
 v is the flow velocity (ft/s)
 n is known as Manning’s n and is a coefficient of roughness
R is the hydraulic radius (a/P) where P is the wetted perimeter (ft)
S is the channel bed slope as a fraction
1.49 is a unit conversion factor. Approximated as 1.5 in the book.
Use 1 if SI (metric) units are used.
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Type of Channel and Description
Table 7.1 Manning’s n Roughness Coefficient
Minimum
Normal
Maximum
Streams
Streams on plain
Clean, straight, full stage, no rifts or deep pools
0.025
0.03
0.033
Clean, winding, some pools, shoals, weeds & stones
0.033
0.045
0.05
Same as above, lower stages and more stones
0.045
0.05
0.06
0.05
0.07
0.07
0.075
0.1
0.15
Bottom: gravels, cobbles, and few boulders
0.03
0.04
0.05
Bottom: cobbles with large boulders
0.04
0.05
0.07
Sluggish reaches, weedy, deep pools
Very weedy reaches, deep pools, or floodways
with heavy stand of timber and underbrush
Mountain streams, no vegetation in channel, banks steep, trees & brush along
banks submerged at high stages
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Channel Conditions
Material Involved
Degree of irregularity
Variations of Channel Cross Section
Earth
Values
no
Rock Cut
0.025
Fine Gravel
0.024
Coarse Gravel
0.027
Smooth
n1
0.005
Moderate
0.010
Severe
0.020
Gradual
n2
Negligible
0.000
0.005
Alternating Frequently
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0.000
Minor
Alternating Occasionally
Relative Effect of Obstructions
0.020
0.010-0.015
n3
0.000
Minor
0.010-0.015
Appreciable
0.020-0.030
Severe
0.040-0.060
Table 7.2. Values for the computation of the roughness coefficient (Chow, 1959)
Channel Conditions
Material Involved
Degree of irregularity
Variations of Channel Cross
Section
Relative Effect of Obstructions
Vegetation
Degree of Meandering
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Earth
Values
n0
0.025
Rock Cut
0.025
Fine Gravel
0.024
Coarse Gravel
0.027
Smooth
n1
0.000
Minor
0.005
Moderate
0.010
Severe
0.020
Gradual
n2
0.000
Alternating Occasionally
0.005
Alternating Frequently
0.010-0.015
Negligible
n3
0.000
Minor
0.010-0.015
Appreciable
0.020-0.030
Severe
0.040-0.060
Low
n4
0.005-0.010
Medium
0.010-0.025
High
0.025-0.050
Very High
0.050-0.100
Minor
m5
1.000
Appreciable
1.150
Severe
1.300
n = (n0 + n1 + n2 + n3 + n4 ) m5
Equation 7.12
Example Problem
Velocity & Discharge
 Channel geometry known
 Depth of flow known
 Determine the flow velocity and discharge
20 ft
1.5 ft
 Bed slope of 0.002 ft/ft
 Manning’s n of 0.04
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Solution
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q = va equation 7.1
v =(1.5/n) R2/3 S1/2 (equation 7.2)
R= a/P (equation 7.3)
a = width x depth = 20 x 1.5 ft = 30 ft2
P= 20 + 1.5 + 1.5 ft = 23 ft.
R= 30/23 = 1.3 ft
S = 0.002 ft/ft (given) and n = 0.04 (given)
v = (1.5/0.04)(1.3)2/3(0.002)1/2 = 2 ft/s
q = va=2x30= 60 ft3/s or 60 cfs
Answer: the velocity is 2 ft/s and the discharge is 60 cfs
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Example Problem
Velocity & Discharge
 Discharge known
 Channel geometry known
 Determine the depth of flow
35 ft
? ft
 Discharge is 200 cfs
 Bed slope of 0.005 ft/ft
 Stream on a plain, clean, winding, some pools and stones
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Table 7.1 Manning’s n Roughness Coefficient
Type of Channel and Description
Minimum
Normal
Maximum
Streams
Streams on plain
Clean, straight, full stage, no rifts or deep pools
0.025
0.03
0.033
Clean, winding, some pools, shoals, weeds & stones
0.033
0.045
0.05
Same as above, lower stages and more stones
0.045
0.05
0.06
0.05
0.07
0.07
0.075
0.1
0.15
Bottom: gravels, cobbles, and few boulders
0.03
0.04
0.05
Bottom: cobbles with large boulders
0.04
0.05
0.07
Sluggish reaches, weedy, deep pools
Very weedy reaches, deep pools, or floodways
with heavy stand of timber and underbrush
Mountain streams, no vegetation in channel, banks steep, trees & brush along
banks submerged at high stages
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Solution
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q = va equation 7.1
v =(1.5/n) R2/3 S1/2 (equation 7.2)
R= a/P (equation 7.3)
Guess a depth! Lets try 2 ft
a = width x depth = 35 x 2 ft = 70 ft 2
P= 35 + 2 + 2 ft = 39 ft.
R= 70/39 = 1.8 ft
S = 0.005 ft/ft (given)
n = 0.033 to 0.05 (Table 7.1) Consider deepest depth
v = (1.5/0.05)(1.8)2/3(0.005)1/2 = 3.1 ft/s
q = va=3.1 x 70= 217 ft3/s or 217 cfs
If the answer is <10% different from the target stop!
Answer: The flow depth is about 2 ft for a discharge of 200 cfs
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Darcy-Weisbach Equation
• Hey’s version of the equation:
8 gRS
v 
f
2
f is the Darcy-Weisbach resistance factor
and all dimensions are in SI units.
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Hey (1979) Estimate Of “f”
• Hey’s version of the equation:
f
0.5
 aR
 2.03
 3.5D84




a is a function of the cross-section and all
dimensions are in SI units.
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Bathurst (1982) Estimate Of “a”
 R 
a  11 .1

dm 
0.314
dm is the maximum depth at the cross-section
provided the width to depth ratio is greater than 2.
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Flow in Compound Channels
Most
flow occurs in main channel; however
during flood events overbank flows may
occur.
In
this case the channel is broken into crosssectional parts and the sum of the flow is
calculated for the various parts.
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Flow in Compound Channels
• Natural channels often have a main channel and an overbank
section.
Overbank Section
Main Channel
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Flow in Compound Channels
1.49 1/ 2  A i 
Vi 
S  
ni
 Pi 
23
n
Q   Vi A i
i 1
In determining R only that part of the wetted perimeter
in contact with an actual channel boundary is used.
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Channel and Floodplain Subdivision
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Variation in Manning’s “n”
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Section Plan
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Shallow Overbank Flow
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Deep Overbank Flow
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