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Pertemuan 17 - 18 Open Channel 1. 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

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## 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.**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.**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 X Wetted Perimeter 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 Area = A = B x Yn Wetted Perimeter = P = B + 2Yn Hydraulic Radius = A/P = R = BYn/(B+2Yn) Yn B**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 = 0.47 meters Yn 2 m**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’ Θ = 20° Concrete Linedn = 0.015 35’ A, P and R for Trapezoidal Channels A = Yn(B + Yn cot θ) P = B + (2Yn/sin θ ) R = (Yn(B + Yn cot θ)) / (B + (2Yn/sin θ )) Yn θ B**The Trapezoidal Channel**Slope S = 0.0003 ft/ft 25’ Θ = 20° Concrete Linedn = 0.015 35’ 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**The Trapezoidal Channel**Slope S = 0.0003 ft/ft 25’ Θ = 20° Concrete Linedn = 0.015 35’ 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**Manning’s Over Different Terrains**S = .005 ft/ft 5’ 5’ 5’ 3’ 3’ Grassn=.03 Grassn=.03 Concrete n=.015 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.**Manning’s Over Different Terrains**S = .005 ft/ft 5’ 5’ 5’ 3’ 3’ Grassn=.03 Grassn=.03 Concrete n=.015 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)1.882/3√(.005) Q = 80.24 cfs per section For both sections… Q = 2 x 80.24 = 160.48 cfs**Manning’s Over Different Terrains**S = .005 ft/ft 5’ 5’ 5’ 3’ 3’ Grassn=.03 Grassn=.03 Concrete n=.015 The Concrete portionsA = 5’ x 6’ = 30 ft2 P = 5’ + 3’ + 3’= 11 ft R = 30 ft2/11 ft = 2.72 ft Q = 30(1.49/.015)2.722/3√(.005)Q = 410.6 cfs For the entire channel… Q = 410.6 + 129.3 = 540 cfs**Uniform Open Channel Flow**• Basic relationships • Continuity equation • Energy equation • Momentum equation • Resistance equations**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**3a**Inflow 3 A Change in Storage 3b Outflow 1 2 A Section AA Continuity Equation Inflow – Outflow = Change in Storage**Area of the cross-section**(ft2) or (m2) Avg. velocity of flow at a cross-section (ft/s) or (m/s) Flow rate (cfs) or (m3/s) General Flow Equation Q = va Equation 7.1**Resistance (velocity) Equations**• Manning’s Equation • Darcy-Weisbach Equation Equation 7.2 Equation 7.6**Velocity Distribution In A Channel**Depth-averaged velocity is above the bed at about 0.4 times the depth**Manning’s Equation**• In 1889 Irish Engineer, Robert Manning presented the formula: 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.**Table 7.2. Values for the computation of the roughness**coefficient (Chow, 1959) n = (n0 + n1 + n2 + n3 + n4 ) m5Equation 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**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**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**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**Darcy-Weisbach Equation**• Hey’s version of the equation: f is the Darcy-Weisbach resistance factor and all dimensions are in SI units.**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.**Bathurst (1982) Estimate Of “a”**dm is the maximum depth at the cross-section provided the width to depth ratio is greater than 2.**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.**Overbank Section**Main Channel Flow in Compound Channels • Natural channels often have a main channel and an overbank section.**Flow in Compound Channels**In determining R only that part of the wetted perimeter in contact with an actual channel boundary is used.

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