Pertemuan 17 — 18 Open Channel 1 Bina

Pertemuan 17 - 18 Open Channel 1 Bina Nusantara

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/ftm/m n or ) = 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 Bina Nusantara Wetted Perimeter X Slope = S = Y/X

Manning’s Equation 2/3 V = (1/n)R √(S) for the metric system 2/3 V = (1. 49/n)R √(S) 2/3 Q = A(k/n)R √(S) for the English system 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 Bina Nusantara Area = A = B x Yn Wetted Perimeter = P = B + 2 Yn Hydraulic Radius = A/P = R = BYn/(B+2 Yn)

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 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 = 2 Yn P = 2 + 2 Yn R = 2 Yn/(2 + 2 Yn) Next, apply Manning’s equation Q = A(1/n)R 2/3√(S) 2. 28 = (2 Yn)x(1/0. 015)x(2 Yn/(2 + 2 Yn))2/3 x√(0. 006) Solving for Y n Yn Bina Nusantara 2 m 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 + (2 Yn/sin θ ) Yn θ B Bina Nusantara R = (Yn(B + Yn cot θ)) / (B + (2 Yn/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) 2 A = 25( 35 + 25 x cot(20)) = 2592 ft P = B + (2 Yn/sin θ ) P = 35 + (2 x 25/sin(20)) = 181. 2 ft R = (Yn(B + Yn cot θ)) / (B + (2 Yn/sin θ )) R = 2592’ / 181. 2’ = 14. 3 ft Bina Nusantara

The Trapezoidal Channel Slope S = 0. 0003 ft/ft 25’ Concrete Lined n = 0. 015 35’ Θ = 20° 2 A = 2592 ft R = 14. 3 ft 2/3 Q = A(1. 49/n)R √(S) 2/3 Q = 2592 x (1. 49 /. 015) x 14. 3 x √(. 0003) Q = Max allowable Flow = 26, 273 cfs Bina Nusantara

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. Bina Nusantara

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 ft 2 For each section: P = 5’ + 3’ = 8 ft R = 15 ft 2/8 ft = 1. 88 ft 2/3 Q = 15(1. 49/. 03)1. 88√(. 005) Q = 80. 24 cfs per section For both sections… Bina Nusantara 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 ft 2 Concrete n=. 015 Grass n=. 03 3’ The Concrete portions P = 5’ + 3’= 11 ft R = 30 ft 2/11 ft = 2. 72 ft 2/3 Q = 30(1. 49/. 015)2. 72√(. 005) Q = 410. 6 cfs For the entire channel… Bina Nusantara Q = 410. 6 + 129. 3 = 540 cfs

Uniform Open Channel Flow Basic relationships Continuity equation Energy equation Momentum equation Resistance equations Bina Nusantara

Flow in Streams Introduction n Effective Discharge n Shear Stresses n Pattern & Profile n ü Open Channel Hydraulics ü Resistance Equations ü Compound Channel • Sediment Transport • Bed Load Movement • Land Use and Land Use Change Bina Nusantara

Continuity Equation Inflow 3 3 a A Change in Storage 3 b Outflow 1 A 2 Section AA Inflow – Outflow = Change in Storage Bina Nusantara

General Flow Equation Q = va Flow rate (cfs) or (m 3/s) Bina Nusantara Avg. velocity of flow at a cross-section (ft/s) or (m/s) Equation 7. 1 Area of the cross-section (ft 2) or (m 2)

Resistance (velocity) Equations §Manning’s Equation 7. 2 §Darcy-Weisbach Equation Bina Nusantara Equation 7. 6

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: Equation 7. 2 ü v is the flow velocity (ft/s) ü n is known as Manning’s n and is a coefficient of roughness üR is the hydraulic radius (a/P) where P is the wetted perimeter (ft) üS is the channel bed slope as a fraction ü 1. 49 is a unit conversion factor. Approximated as 1. 5 in the book. Use 1 if SI (metric) units are used. Bina Nusantara

Type of Channel and Description Table 7. 1 Manning’s n Roughness Coefficient Minimum Normal Maximum Streams on plain Clean, straight, full stage, no rifts or deep pools 0. 025 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. 06 0. 05 0. 075 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 Bina Nusantara

Channel Conditions Material Involved Earth Values no 0. 020 Rock Cut Fine Gravel 0. 024 Coarse Gravel Degree of irregularity 0. 025 0. 027 Smooth n 1 0. 000 Minor Moderate 0. 010 Severe Variations of Channel Cross Section 0. 005 0. 020 Gradual n 2 Alternating Occasionally 0. 005 Alternating Frequently Relative Effect of Obstructions Negligible 0. 000 0. 010 -0. 015 n 3 0. 000 Minor 0. 010 -0. 015 Appreciable 0. 020 -0. 030 Severe 0. 040 -0. 060 Bina Nusantara Vegetation Low n 4 0. 005 -0. 010

Table 7. 2. Values for the computation of the roughness coefficient (Chow, 1959) Bina Nusantara n = (n 0 + n 1 + n 2 + n 3 + n 4 ) m 5 Equation 7. 12

Example Problem Velocity & Discharge ü Channel geometry known ü Depth of flow known ü Determine the flow velocity and discharge 20 ft 1. 5 ft ü Bed slope of 0. 002 ft/ft ü 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= 20 + 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=2 x 30= 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 ü Discharge known ü Channel geometry known ü Determine the depth of flow 35 ft ? ft ü Discharge is 200 cfs ü Bed slope of 0. 005 ft/ft ü Stream on a plain, clean, winding, some pools and stones Bina Nusantara

Table 7. 1 Manning’s n Roughness Coefficient Type of Channel and Description Minimum Normal Maximum Streams on plain Clean, straight, full stage, no rifts or deep pools 0. 025 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. 06 0. 05 0. 075 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 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= 35 + 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 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” dm is the maximum depth at the cross-section provided the width to depth ratio is greater than 2. Bina Nusantara

Flow in Compound Channels n. Most flow occurs in main channel; however during flood events overbank flows may occur. n. In this case the channel is broken into crosssectional 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. Overbank Section Main Channel 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 Bina Nusantara