Pertemuan Open Channel 1

Bina Nusantara

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.

Bina Nusantara 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.

Bina Nusantara 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 = for concrete n = 0.03 for clean natural channel n =.01 for glass Y n = Normal depth (depth of uniform flow) Area Wetted Perimeter YnYn Y X Slope = S = Y/X

Bina Nusantara Manning’s Equation V = (1/n)R 2/3 √(S) for the metric system V = (1.49/n)R 2/3 √(S) for the English system Q = A(k/n)R 2/3 √(S) k is either 1 or 1.49 As you can see, Y n is not directly a part of Manning’s equation. However, A and R depend on Y n. 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 Y n Wetted Perimeter = P = B + 2Y n Hydraulic Radius = A/P = R = BY n /(B+2Y n ) B YnYn

Bina Nusantara Simple Manning’s Example A rectangular open concrete (n=0.015) channel is to be designed to carry a flow of 2.28 m 3 /s. The slope is m/m and the bottom width of the channel is 2 meters. Determine the normal depth that will occur in this channel. 2 m YnYn First, find A, P and R A = 2Y n P = 2 + 2Y n R = 2Y n /(2 + 2Y n ) Next, apply Manning’s equation Q = A(1/n)R 2/3 √(S)  2.28 = (2Y n )x(1/0.015)x(2Y n /(2 + 2Yn)) 2/3 x√(0.006) Solving for Y n Y n = 0.47 meters

Bina Nusantara 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? 25’ 35’ Θ = 20° Concrete Lined n = Slope S = ft/ft A, P and R for Trapezoidal Channels B YnYn θ A = Y n (B + Y n cot θ) P = B + (2Y n /sin θ ) R = (Y n (B + Y n cot θ)) / (B + (2Y n /sin θ ))

Bina Nusantara The Trapezoidal Channel 25’ 35’ Θ = 20° Concrete Lined n = Slope S = ft/ft A = Y n (B + Y n cot θ) A = 25( x cot(20)) = 2592 ft 2 P = B + (2Y n /sin θ ) P = 35 + (2 x 25/sin(20)) = ft R = (Y n (B + Y n cot θ)) / (B + (2Y n /sin θ )) R = 2592’ / 181.2’ = 14.3 ft

Bina Nusantara The Trapezoidal Channel 25’ 35’ Θ = 20° Concrete Lined n = Slope S = ft/ft A = 2592 ft 2 R = 14.3 ft Q = A(1.49/n)R 2/3 √(S) Q = 2592 x (1.49 /.015) x /3 x √(.0003) Q = Max allowable Flow = 26,273 cfs

Bina Nusantara Manning’s Over Different Terrains 3’ 5’ Grass n=.03 Concrete n=.015 Grass n=.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’ 5’ Grass n=.03 Concrete n=.015 Grass n=.03 The Grassy portions: For each section: A = 5’ x 3’ = 15 ft 2 P = 5’ + 3’ = 8 ft R = 15 ft 2 /8 ft = 1.88 ft Q = 15(1.49/.03)1.88 2/3 √(.005) Q = 15(1.49/.03)1.88 2/3 √(.005) Q = cfs per section  For both sections… Q = 2 x = cfs S =.005 ft/ft

Bina Nusantara Manning’s Over Different Terrains 3’ 5’ Grass n=.03 Concrete n=.015 Grass n=.03 The Concrete portions A = 5’ x 6’ = 30 ft 2 P = 5’ + 3’ + 3’= 11 ft R = 30 ft 2 /11 ft = 2.72 ft Q = 30(1.49/.015)2.72 2/3 √(.005) Q = 30(1.49/.015)2.72 2/3 √(.005) Q = cfs For the entire channel… Q = = 540 cfs S =.005 ft/ft

Bina Nusantara Uniform Open Channel Flow Basic relationships Continuity equation Energy equation Momentum equation Resistance 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 (m 3 /s) Avg. velocity of flow at a cross-section (ft/s) or (m/s) Area of the cross-section (ft 2 ) or (m 2 ) 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: 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 DescriptionMinimumNormalMaximum Streams Streams on plain Clean, straight, full stage, no rifts or deep pools Clean, winding, some pools, shoals, weeds & stones Same as above, lower stages and more stones 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 Bottom: gravels, cobbles, and few boulders Bottom: cobbles with large boulders Table 7.1 Manning’s n Roughness Coefficient

Bina Nusantara Channel ConditionsValues Material InvolvedEarthnono0.020 Rock Cut0.025 Fine Gravel0.024 Coarse Gravel0.027 Degree of irregularitySmoothn Minor0.005 Moderate0.010 Severe0.020 Variations of Channel Cross SectionGradualn Alternating Occasionally0.005 Alternating Frequently Relative Effect of ObstructionsNegligiblen Minor Appreciable Severe VegetationLown Medium High Very High Degree of MeanderingMinorm Appreciable1.150 Severe1.300

Bina Nusantara Table 7.2. Values for the computation of the roughness coefficient (Chow, 1959) n = (n 0 + n 1 + n 2 + n 3 + n 4 ) m 5 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 ft/ft Manning’s n of 0.04

Bina Nusantara Solution q = va equation 7.1 v =(1.5/n) R 2/3 S 1/2 (equation 7.2) R= a/P (equation 7.3) a = width x depth = 20 x 1.5 ft = 30 ft 2 P= ft = 23 ft. R= 30/23 = 1.3 ft S = 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 ft 3 /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 ft/ft Stream on a plain, clean, winding, some pools and stones

Bina Nusantara Type of Channel and DescriptionMinimumNormalMaximum Streams Streams on plain Clean, straight, full stage, no rifts or deep pools Clean, winding, some pools, shoals, weeds & stones Same as above, lower stages and more stones 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 Bottom: gravels, cobbles, and few boulders Bottom: cobbles with large boulders Table 7.1 Manning’s n Roughness Coefficient

Bina Nusantara Solution q = va equation 7.1 v =(1.5/n) R 2/3 S 1/2 (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= ft = 39 ft. R= 70/39 = 1.8 ft S = ft/ft (given) n = 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 ft 3 /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.

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.

Bina Nusantara Bathurst (1982) Estimate Of “a” d m is the maximum depth at the cross-section provided the width to depth ratio is greater than 2.

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 Channels 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