Chapter 9. Steady Flow in Open channels


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1 Chapter 9 Steady Flow in Open channels
2 Objectives Be able to define uniform open channel flow Solve uniform open channel flow using the Manning Equation
3 9.1 Uniform Flow in Open Channel Openchannel flows are characterized by the presence of a liquidgas interface called the free surface. Natural flows: rivers, creeks, floods, etc. Humanmade systems: freshwater aqueducts, irrigation, sewers, drainage ditches, etc.
4 In an open channel, Velocity is zero on bottom and sides of channel due to noslip condition Velocity is maximum at the midplane of the free surface In most cases, velocity also varies in the streamwise direction Therefore, the flow is 3D Nevertheless, 1D approximation is made with good success for many practical problems.
5 differences between pipe flow and open channel flow The flow of water in a conduit may be either open channel flow or pipe flow. The two kinds of flow are similar in many ways but differ in one important respect. Openchannel flow must have a free surface, whereas pipe flow has none. A free surface is subject to atmospheric pressure. In Pipe flow there exist no direct atmospheric flow but hydraulic pressure only.
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7 Open channel flow is driven by gravity rather than by pressure work as in pipes.
8 9.2 Classification of OpenChannel Flows The most common classification method is by rate of change of freesurface depth. The classes are summarized as 1. Uniform flow (constant depth and slope) 2. Varied flow a. Gradually varied (onedimensional) b. Rapidly varied (multidimensional)
9 Classification of OpenChannel Flows Flow in open channels is also classified as being uniform or nonuniform, depending upon the depth y. Uniform flow (UF) encountered in long straight sections where head loss due to friction is balanced by elevation drop. Depth in UF is called normal depth y n
10 Classification of OpenChannel Flows Obstructions cause the flow depth to vary. Rapidly varied flow (RVF) occurs over a short distance near the obstacle. Gradually varied flow (GVF) occurs over larger distances and usually connects UF and RVF.
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12 9.3 Properties of open channels Artificial channels These are channels made by man. They include irrigation canals, navigation canals, spillways, sewers, culverts and drainage ditches. They are usually constructed in a regular crosssection shape throughout and are thus prismatic channels (they don t widen or get narrower along the channel. Natural channels Natural channels can be very different. They are not regular nor prismatic and their materials of construction can vary widely (although they are mainly of earth this can possess many different properties.) The surface roughness will often change with time distance and even elevation.
13 Properties of open channels Geometric properties necessary for analysis Depth (y) the vertical distance from the lowest point of the channel section to the free surface. Stage (z) the vertical distance from the free surface to an arbitrary datum Area (A) the crosssectional area of flow, normal to the direction of flow Wetted perimeter (P) the length of the wetted surface measured normal to the direction of flow. Surface width (B) width of the channel section at the free surface Hydraulic radius (R) the ratio of area to wetted perimeter (A/P) Hydraulic mean depth (Dm) the ratio of area to surface width (A/B)
14 The wetted perimeter does not include the free surface.
15 Multiple Choice Consider an open rectangular channel 3m wide laid on a 1 slope. If the water depth is 2m, the hydraulic radius is: (a) 0.43m (b) 0.60m (c) 0.86m (d) 1.00m
16 9.4 Continuity and Energy Equations 1D steady continuity equation can be expressed as 1D steady energy equation between two stations Head loss h L is expressed as 2 L V h L = f 4R 2g
17 9.5 Uniform Flow in Channels Occurs in long straight runs of constant slope The velocity is constant with V = V o Water depth is constant with y = y n Slope is constant with S o = tanα
18 Head loss in a Channel Recall: DarcyWeisbach Equation For circular pipes: L V h = f f D 2g 2 For noncircular conduits: 2 L V h = f R = f 4R 2g A χ
19 Head loss in a Channel 2 L V h f = S L = f 0 4R 2g Solve for V: V = 4S 0 R2g f V = 8g f R 1/ 2 S 1/ 2 0
20 Chezy Formula For a given channel shape and roughness: 8g = Constant f This leads to the Chezy Formula = C Named after Antoine Chezy, who did experiments in the River Seine in 1760s V C RS = = 0 C Ri C Chezy Coefficient i the bed slope of the channel
21 Manning s Analysis Robert Manning (Irish engineer, 1880s) Found C increases with channel size: C = 1 R 1/ 6 n Manning s formula Where n = a roughness coefficient
22 Values of Manning n Lined Canals n Cement plaster Untreated gunite Wood, planed Wood, unplaned Concrete, trowled Concrete, wood forms, unfinished Rubble in cement Asphalt, smooth Asphalt, rough Natural Channels Gravel beds, straight Gravel beds plus large boulders Earth, straight, with some grass Earth, winding, no vegetation Earth, winding with vegetation 0.050
23 How to get n? Field studies expensive! References Estimate based on roughness height
24 9.6 Computations in uniform flow We can use Manning's formula for discharge to calculate steady uniform flow. Two calculations are usually performed to solve uniform flow problems. 1. Discharge from a given depth 2. Depth for a given discharge In steady uniform flow the flow depth is know as normal depth.
25 example 1 Discharge from depth in a trapezoidal channel A concrete lined trapezoidal channel with uniform flow has a normal depth is 2m.The base width is 5m and the side slopes are equal at 1:2 Manning's n can be taken as And the bed slope S0 = What are: a) Discharge (Q) b) Mean velocity (V) c) Reynolds number (Re)
26 Solution: Calculate the section properties the mean velocity And the Reynolds number This is very large  i.e. well into the turbulent zone  the application of the Manning's equation was therefore valid.
27 Example: Manning Formula What is the flow capacity of a finished concrete channel that drops 1.2 m in 3 km? m 3 m
28 Example solution 1 Q = AR n 2/3 S 1/ 2 0 A = ( 3m)(1.5m) + (3m)(1.5m) = 9m P = = ( 3m) + 2 (3m) + (1.5m) 9.71m n = S 1 Q = (9m Q = 14.3m / s m = = m )(0.927m) 2/3 2 (0.0004) 1/ 2 A R = = m P
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31 9.7 Best Hydraulic Cross Sections Best hydraulic cross section for an open channel is the one with the minimum wetted perimeter for a specified cross section (or maximum hydraulic radius R h ) Also reflects economy of building structure with smallest perimeter
32 Best Hydraulic Cross Sections Example: Rectangular Channel Cross section area, A c = yb Perimeter, p = b + 2y Solve A c for b and substitute Taking derivative with respect to To find minimum, set derivative to zero Best rectangular channel has a depth 1/2 of the width
33 Best Hydraulic Cross Sections Same analysis can be performed for a trapezoidal channel Similarly, taking the derivative of p with respect to q, shows that the optimum angle is For this angle, the best flow depth is
34 Multiple Choice Consider an open rectangular channel 3m wide laid on a 1 slope. The most efficient water depth (best depth for a given flow and resistance ) is: (a) 1.0m (b) 1.5m (c) 2.0m (d) 2.5m
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