Pertemuan 17 - 18Open Channel 1

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Bina Nusantara

Open Channel FlowUniform 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.

Bina Nusantara

Open Channel FlowManning’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.

Bina Nusantara

Manning’s EquationTerms 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

Wetted Perimeter

Yn

Y

XSlope = S = Y/X

Bina Nusantara

Manning’s Equation

V = (1/n)RV = (1/n)R2/32/3√√(S)(S) for the metric systemfor the metric system

V = (1.49/n)RV = (1.49/n)R2/32/3√√(S)(S) for the English systemfor the English system

Q = A(k/n)RQ = A(k/n)R2/32/3√√(S)(S) k is either 1 or 1.49k 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

Area = A = B x Yn

Wetted Perimeter = P = B + 2Yn

Hydraulic Radius = A/P = R = BYn/(B+2Yn)

B

Yn

Bina Nusantara

Simple Manning’s ExampleA 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.

2 m

Yn

First, find A, P and R

A = 2Yn P = 2 + 2Yn R = 2Yn/(2 + 2Yn)

Next, apply Manning’s equation

Q = A(1/n)RQ = A(1/n)R2/32/3√(S) √(S)

2.28 = (2Y2.28 = (2Ynn)x(1/0.015)x(2Y)x(1/0.015)x(2Ynn/(2 + 2Yn))/(2 + 2Yn))2/32/3x√(0.006)x√(0.006)

Solving for Yn

Yn = 0.47 meters

Bina Nusantara

The Trapezoidal ChannelHouse flooding occurs along Brays Bayou when

water overtops the banks. What flow is allowable in Brays Bayou if it has the geometry shown below?

25’

35’

Θ = 20°Concrete Lined

n = 0.015

Slope S = 0.001

ft/ft

A, P and R for Trapezoidal Channels

B

Yn

θ

A = Yn(B + Yn cot θ)

P = B + (2Yn/sin θ )

R = (Yn(B + Yn cot θ)) / (B + (2Yn/sin θ ))

Bina Nusantara

The Trapezoidal Channel

25’

35’

Θ = 20°Concrete Lined

n = 0.015

Slope S = 0.0003 ft/ft

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

Bina Nusantara

The Trapezoidal Channel

25’

35’

Θ = 20°Concrete Lined

n = 0.015

Slope S = 0.0003

ft/ft

A = 2592 ftA = 2592 ft22 R = 14.3 ft R = 14.3 ft

Q = A(1.49/n)RQ = A(1.49/n)R2/32/3√√(S)(S)Q = 2592 x (1.49 / .015) x 14.3Q = 2592 x (1.49 / .015) x 14.32/32/3 x √(.0003) x √(.0003)

Q = Max allowable Flow = 26,273 cfsQ = Max allowable Flow = 26,273 cfs

Bina Nusantara

Manning’s Over Different Terrains

3’

3’

5’ 5’ 5’

Grassn=.03 Concrete

n=.015

Grassn=.03

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.

S = .005 ft/ft

Bina Nusantara

Manning’s Over Different Terrains

3’

3’

5’ 5’ 5’

Grassn=.03 Concrete

n=.015

Grassn=.03

The Grassy portions:

For each section: A = 5’ x 3’ = 15 ft2 P = 5’ + 3’ = 8 ft R = 15 ft2/8 ft

= 1.88 ft

Q = 15(1.49/.03)1.88Q = 15(1.49/.03)1.882/32/3√√(.005)(.005) Q = 80.24 cfs per section For both sections…

Q = 2 x 80.24 = 160.48 cfs

S = .005 ft/ft

Bina Nusantara

Manning’s Over Different Terrains

3’

3’

5’ 5’ 5’

Grassn=.03 Concrete

n=.015

Grassn=.03

The Concrete portions A = 5’ x 6’ = 30 ft2 P = 5’ + 3’ + 3’= 11 ft R =

30 ft2/11 ft = 2.72 ft

Q = 30(1.49/.015)2.72Q = 30(1.49/.015)2.722/32/3√√(.005)(.005) Q = 410.6 cfs

For the entire channel…

Q = 410.6 + 129.3 = 540 cfs

S = .005 ft/ft

Bina Nusantara

Uniform Open Channel Flow

Basic relationshipsContinuity equationEnergy equationMomentum equationResistance equations

Bina Nusantara

Flow in Streams

Open Channel Hydraulics Resistance Equations Compound Channel

Introduction Effective Discharge Shear Stresses Pattern & Profile

• Sediment Transport • Bed Load Movement• Land Use and Land Use Change

Bina Nusantara

Continuity Equation

Inflow – Outflow = Change in Storage

Inflow

1 2

A

A

3

Section AA

Change in Storage

Outflow

3a

3b

Bina Nusantara

General Flow Equation

Q = va

Flow rate (cfs) or (m3/s)

Avg. velocity of flow at a cross-section (ft/s) or (m/s)

Area of the cross-section

(ft2) or (m2)

Equation 7.1

Bina Nusantara

Resistance (velocity) Equations

Manning’s Equation

Darcy-Weisbach Equation

Equation 7.2

Equation 7.6

Bina Nusantara

Velocity Distribution In A Channel

Depth-averaged velocity is above the bed at about 0.4 times the depth

Bina Nusantara

Manning’s Equation

• In 1889 Irish Engineer, Robert Manning presented the formula:

2132 SRn

49.1v

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.

Equation 7.2

Bina Nusantara

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

Sluggish reaches, weedy, deep pools 0.05 0.07 0.07

Very weedy reaches, deep pools, or floodways 0.075 0.1 0.15

with heavy stand of timber and underbrush

Mountain streams, no vegetation in channel, banks steep, trees & brush along banks submerged at high stages

Bottom: gravels, cobbles, and few boulders 0.03 0.04 0.05

Bottom: cobbles with large boulders 0.04 0.05 0.07

Table 7.1 Manning’s n Roughness Coefficient

Bina Nusantara

Channel Conditions Values

Material Involved

Earth no 0.020

Rock Cut 0.025

Fine Gravel 0.024

Coarse Gravel 0.027

Degree of irregularity Smooth n1 0.000

Minor 0.005

Moderate 0.010

Severe 0.020

Variations of Channel Cross

SectionGradual n2 0.000

Alternating Occasionally 0.005

Alternating Frequently 0.010-0.015

Relative Effect of

ObstructionsNegligible n3 0.000

Minor 0.010-0.015

Appreciable 0.020-0.030

Severe 0.040-0.060

Vegetation Low n4 0.005-0.010

Medium 0.010-0.025

High 0.025-0.050

Very High 0.050-0.100

Degree of Meandering Minor m5 1.000

Appreciable 1.150

Severe 1.300

Bina Nusantara

Channel Conditions Values Material Involved Earth n0 0.025

Rock Cut 0.025 Fine Gravel 0.024 Coarse Gravel 0.027 Degree of irregularity Smooth n1 0.000 Minor 0.005 Moderate 0.010 Severe 0.020 Variations of Channel Cross

Section Gradual n2 0.000

Alternating Occasionally 0.005 Alternating Frequently 0.010-0.015 Relative Effect of Obstructions Negligible n3 0.000 Minor 0.010-0.015 Appreciable 0.020-0.030 Severe 0.040-0.060 Vegetation Low n4 0.005-0.010 Medium 0.010-0.025 High 0.025-0.050 Very High 0.050-0.100 Degree of Meandering Minor m5 1.000 Appreciable 1.150 Severe 1.300

Table 7.2. Values for the computation of the roughness coefficient (Chow, 1959)

n = (n0 + n1 + n2 + n3 + n4 ) m5 Equation 7.12

Bina Nusantara

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

Bina Nusantara

Solution

• 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

Bina Nusantara

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

Bina Nusantara

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

Sluggish reaches, weedy, deep pools 0.05 0.07 0.07

Very weedy reaches, deep pools, or floodways 0.075 0.1 0.15

with heavy stand of timber and underbrush

Mountain streams, no vegetation in channel, banks steep, trees & brush along banks submerged at high stages

Bottom: gravels, cobbles, and few boulders 0.03 0.04 0.05

Bottom: cobbles with large boulders 0.04 0.05 0.07

Table 7.1 Manning’s n Roughness Coefficient

Bina Nusantara

Solution

• 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 ft2

• 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

Bina Nusantara

Darcy-Weisbach Equation• Hey’s version of the equation:

f is the Darcy-Weisbach resistance factor and all dimensions are in SI units.

f

gRSv

82

Bina Nusantara

Hey (1979) Estimate Of “f”

• Hey’s version of the equation:

a is a function of the cross-section and all dimensions are in SI units.

84

5.0

5.303.2

D

aRf

Bina Nusantara

Bathurst (1982) Estimate Of “a”

dm is the maximum depth at the cross-section provided the width to depth ratio is greater than 2.

314.0

1.11

md

Ra

Bina Nusantara

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 cross-sectional parts and the sum of the flow is calculated for the various parts.

Bina Nusantara

Flow in Compound Channels• Natural channels often have a main channel and

an overbank section.

Main Channel

Overbank Section

Bina Nusantara

Flow in Compound Channels32

i

i2/1

ii P

AS

n

49.1V

i

n

1iiAVQ

In determining R only that part of the wetted perimeter in contact with an actual channel boundary is used.

Bina Nusantara

Channel and Floodplain Subdivision

Bina Nusantara

Variation in Manning’s “n”

Bina Nusantara

Section Plan

Bina Nusantara

Shallow Overbank Flow

Bina Nusantara

Deep Overbank Flow