CVE 372 HYDROMECHANICS OPEN CHANNEL FLOW II

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1 CVE 372 HYDROMECHANICS OPEN CHANNEL FLOW II Dr. Bertuğ Akıntuğ Department of Civil Engineering Middle East Technical University Northern Cyprus Campus CVE 372 Hydromechanics 1/68

2 Overview 3.4 Rapidly Varied Flow Specific Force and Conjugate Depth Hydraulic Jump 3.5 Gradually Varied Flow General Equation of Gradually Varied Flow Types of Slopes Longitudinal Flow Profiles 3.6 Design of Open Channels Hydraulic Efficiency of Cross-sections Design of Nonerodible Channels Design of Erodible Channels CVE 372 Hydromechanics 2/68

3 Rapidly Varied Flow Specific Force and Conjugate Depth Specific force for any channel section is given by Specific Force F = ya + 2 Q ga CVE 372 Hydromechanics 3/68

4 Rapidly Varied Flow Specific Force and Conjugate Depth F = ya + 2 Q ga Specific Energy Curve Channel Cross-Section Specific Force Curve At Critical Depth: df dy = 0, da Q T = T, F =1, and r = 1, 3 dy ga 2 CVE 372 Hydromechanics 4/68

5 Rapidly Varied Flow Specific Force and Conjugate Depth Specific force becomes minimum when the flow is critical. F = ya + 2 Q ga Lower limb AC corresponds to supercritical flow. Upper limb BC corresponds to subcritical flow. For a given F and Q, there are two possible flow regimes represented by the depths y 1 and y 2 which are called conjugate or sequent depths. For rectangular channels, the specific force can be written as specific force per unit width as F b = 1 2 y q gy CVE 372 Hydromechanics 5/68

6 Rapidly Varied Flow Hydraulic Jump Hydraulic jump is a rapidly varied flow in which flow changes abruptly from supercritical flow to a subcritical flow accompanied by considerable turbulence and energy loss. F = F E E + 1 = 2 h L CVE 372 Hydromechanics 6/68 y 1 y A Q ga 1 1 = y 2 2 A Q ga For rectangular channels: y ( ) 1+ 8F = 2 r1 1 2 h L = ( y y ) 2 4y y

7 Rapidly Varied Flow Hydraulic Jump For a simple jump, since F 1 =F 2, y 1 and y 2 are conjugate (sequent) depths. CVE 372 Hydromechanics 7/68

8 Rapidly Varied Flow Hydraulic Jump Practical Applications of Hydraulic Jump To dissipate energy To recover head or rise the water level To increase weight on apron To mix chemicals used for water purification To aerate water Types of Hydraulic Jumps: CVE 372 Hydromechanics 8/68

9 3.4.2 Hydraulic Jump Rapidly Varied Flow CVE 372 Hydromechanics 9/68

10 3.4.2 Hydraulic Jump Rapidly Varied Flow CVE 372 Hydromechanics 10/68

11 3.4.2 Hydraulic Jump Rapidly Varied Flow CVE 372 Hydromechanics 11/68

12 Rapidly Varied Flow Hydraulic Jump Example 9: Determine the type of hydraulic jump for the information given in the figure below. T y 2 =1.5 m y 1 Q=? m 3 /s y 1 =0.8 m 1 (1) (2) CVE 372 Hydromechanics 12/68

13 Rapidly Varied Flow Hydraulic Jump Example 10: A hydraulic jump take place in a rectangular channel of 10 m wide. The initial depth and the velocity of the jump are 0.6 m and m/s, respectively. If the depth of flow on the step is 2.44 m, compute a) the step height, z b) the force on the step. y 3 =2.44 m b = 10 m F b V 1 =10.67 m/s y 1 =0.6 m z (1) (2) (3) CVE 372 Hydromechanics 13/68

14 Rapidly Varied Flow Hydraulic Jump Example 11: Water is flowing in a rectangular channel of width of 1 m shown in the figure below. a) compute y 2, z, and y 4. b) express the alternative and conjugate depths c) find the force exerted by the flow on the step by writing the momentum equations between section (1) and (3) and (1) and (2), and discuss the reason for different answer. y 1 =1.5 m y 2 =? y 4 =? z=? (1) (2) (3) y 3 =0.3 m (4) b = 1m CVE 372 Hydromechanics 14/68

15 Overview 3.4 Rapidly Varied Flow Specific Force and Conjugate Depth Hydraulic Jump 3.5 Gradually Varied Flow General Equation of Gradually Varied Flow Types of Slopes Longitudinal Flow Profiles 3.6 Design of Open Channels Hydraulic Efficiency of Cross-sections Design of Nonerodible Channels Design of Erodible Channels CVE 372 Hydromechanics 15/68

16 Gradually Varied Flow General Equation of Gradually Varied Flow Only y, z, and v changes along x Q = 0 x (any property) t = 0 CVE 372 Hydromechanics 16/68

17 Gradually Varied Flow General Equation of Gradually Varied Flow CVE 372 Hydromechanics 17/68

18 Gradually Varied Flow General Equation of Gradually Varied Flow CVE 372 Hydromechanics 18/68

19 Gradually Varied Flow Types of Slopes CVE 372 Hydromechanics 19/68

20 Gradually Varied Flow Types of Slopes CVE 372 Hydromechanics 20/68

21 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 21/68

22 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 22/68

23 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 23/68

24 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 24/68

25 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 25/68

26 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 26/68

27 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 27/68

28 Gradually Varied Flow Longitudinal Flow Profile CVE 372 Hydromechanics 28/68

29 Gradually Varied Flow Longitudinal Flow Profile Solution of GVF Equations 1. Direct Integration Method 2. Graphical Integration Method 3. Numerical Step Methods 3.1. The Direct Step Method 3.2. The Standard Step Method CVE 372 Hydromechanics 29/68

30 Gradually Varied Flow Longitudinal Flow Profile The Direct Step Method Consider a short channel reach of length x. Equating the total heads at section 1 and section 2 S 0 x + E1 = E2 + S f x CVE 372 Hydromechanics 30/68

31 Gradually Varied Flow Longitudinal Flow Profile The Direct Step Method CVE 372 Hydromechanics 31/68

32 Rapidly Varied Flow Longitudinal Flow Profile Example 12: Determine the water surface profile for the following rectangular channel with b = 3 mater, n=0.0125, and Q=10 m 3 /s S 0 = b = 3 m S 0 = CVE 372 Hydromechanics 32/68

33 Gradually Varied Flow Longitudinal Flow Profile Example 13: A straight prismatic irrigation channel has a trapezoidal cross-section with bottom width b = 6 m and equal side slopes of 2 horizontal and 1 vertical. The bottom slope and the Manning s roughness coefficient are S 0 = and n = 0.014, respectively. At a certain point regulation works produce a depth of 1.80 m in the channel when the discharge is 9.81 m 3 /s. Using the direct step method find the distance between y = 1.80 m and y= 1.50 m in three steps. CVE 372 Hydromechanics 33/68

34 Gradually Varied Flow Longitudinal Flow Profile Example 14: A reservoir is emptied through a sluice gate yielding a smallest depth of 0.40 m at vena-contracta. The channel bottom slope is S 0 = with n = 0.02 while q = m 3 /s/m. Determine the water depth at a point 60 m downstream (horizontally) of vena-contracta in two steps of equal distances by the standard step method. Assume a wide rectangular channel and neglect the eddy lossess. CVE 372 Hydromechanics 34/68

35 Overview 3.4 Rapidly Varied Flow Specific Force and Conjugate Depth Hydraulic Jump 3.5 Gradually Varied Flow General Equation of Gradually Varied Flow Types of Slopes Longitudinal Flow Profiles 3.6 Design of Open Channels Hydraulic Efficiency of Cross-sections Design of Nonerodible Channels Design of Erodible Channels CVE 372 Hydromechanics 35/68

36 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 36/68

37 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections Example 15: Between two open channel having rectangular and trapezoidal sections as shown, determine which section is hydraulically better. Assume both channels have the same roughness and slope. CVE 372 Hydromechanics 37/68

38 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 38/68

39 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 39/68

40 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 40/68

41 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 41/68

42 Design of Open Channels Hydraulic Efficiency of Cross-sections Recommended Channel Cross-sections CVE 372 Hydromechanics 42/68

43 Design of Open Channels Hydraulic Efficiency of Cross-sections Recommended Channel Cross-sections CVE 372 Hydromechanics 43/68

44 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 44/68

45 Design of Open Channels 2. OPEN CHANNEL FLOW Hydraulic Efficiency of Cross-sections CVE 372 Hydromechanics 45/68

46 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 46/68

47 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 47/68

48 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 48/68

49 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 49/68

50 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 50/68

51 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 51/68

52 Design of Open Channels Design of Nonerodible Channels 2. OPEN CHANNEL FLOW CVE 372 Hydromechanics 52/68

53 Design of Open Channels 2. OPEN CHANNEL FLOW Design of Nonerodible Channels CVE 372 Hydromechanics 53/68

54 Design of Open Channels Design of Nonerodible Channels Example 16 A trapezoidal channel carrying 11.5 m 3 /s clear water is build with nonerodible (concrete) channel having a slope of and n= Proportion the section dimensions. Example 17 Design the channel given in Example 16 by using experience curve and take z=1.5. CVE 372 Hydromechanics 54/68

55 Design of Open Channels Design of Erodible Channels Channels used as irrigation and drainage canals, are unlined for cost reasons and they must be so proportioned as to permit neither silting nor scouring in objectionable quantity. There are two methods of approach to the proper design of erodible channels: Method of maximum permissible velocity Method of permissible tractive force CVE 372 Hydromechanics 55/68

56 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Triactive Force Average unit tractive force For a uniform flow the average unit tractive force acting on the wetted perimeter of the channel section: τ o = γrs o CVE 372 Hydromechanics 56/68

57 Design of Open Channels 2. OPEN CHANNEL FLOW Design of Erodible Channels Method of Maximum Permissible Triactive Force Unit tractive force A typical distribution of unit tractive force in a trapezoidal channel is shown below. Distribution of unit tractive force in a trapezoidal channel with b=4y CVE 372 Hydromechanics 57/68

58 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Tractive Force Maximum unit tractive force 2. OPEN CHANNEL FLOW Following figures show the maximum unit tractive forces on the sides and bottom of various channel sections. Figure 3: Maximum unit tractive forces in terms of γys 0 CVE 372 Hydromechanics 58/68

59 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Triactive Force Permissible tractive force The permissible tractive force is the maximum tractive force that will not cause serious erosion of the material forming the channel bed on a level surface. This unit tractive force is determined by laboratory experiments and the value thus obtained is known as the critical tractive force. CVE 372 Hydromechanics 59/68

60 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Tractive Force Permissible tractive force 2. OPEN CHANNEL FLOW For noncohesive materials the permissible tractive force is a function of the average particle diameter. Figure 4: Unit tractive forces for canals in noncohesive materials CVE 372 Hydromechanics 60/68

61 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Triactive Force Permissible tractive force For cohesive materials triactive force is function of the void ratio 2. OPEN CHANNEL FLOW Figure 5: Unit tractive forces for canals in cohesive materials CVE 372 Hydromechanics 61/68

62 Design of Open Channels 2. OPEN CHANNEL FLOW Design of Erodible Channels Method of Maximum Permissible Triactive Force Motion of the particle on the side aτ s = the tractive force on the side slope W s.sinφ = the gravity force component where a: the effective area of the particle, τ : the unit tractive force on the side of the channel, W s : the submerged weight of the particle. 2 2 [( W sinφ ) + ( aτ ) ] 1/ 2 s s : the resultant force W s cosφ tanθ = the resistance to motion of the particle tanθ = the coefficient of friction (θ is the angle of repose) CVE 372 Hydromechanics 62/68

63 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Triactive Force Motion of the particle on the side (continued) 2. OPEN CHANNEL FLOW W s cos φ tanθ = W 2 s sin 2 φ + a 2 τ 2 s W s tan τ = cos φ tanθ 1 s 2 a tan 2 φ θ CVE 372 Hydromechanics 63/68

64 Design of Open Channels Design of Erodible Channels Method of Maximum Permissible Triactive Force Motion of the particle on the bed W tanθ = s aτ, b τ = unit tractive force on the bottom. b 2. OPEN CHANNEL FLOW W s τ b = tanθ a The ratio of τ s to τ b is called the tractive force ratio. K τ s = τ b = cosφ 1 tan tan 2 2 φ θ K = 1 sin sin 2 2 φ θ CVE 372 Hydromechanics 64/68

65 Design of Open Channels Design of Erodible Channels Design Procedure: Given: Design discharge Q, type of soil and channel bed slope So; 1. Side slope z, roughness coefficient n, and angle of repose θ are selected from the figure. 2. OPEN CHANNEL FLOW Figure 6: Angle of repose of noncohesive materials CVE 372 Hydromechanics 65/68

66 Design of Open Channels 2. OPEN CHANNEL FLOW Design of Erodible Channels Design Procedure: 2. The permissible tractive force,τ p, is determined according to cohessiveness (Noncohesive or Cohesive) CVE 372 Hydromechanics 66/68

67 Design of Open Channels Design of Erodible Channels Design Procedure: 3. Any b/y ratio is assumed and tractive force of water; On the channel bed, C 1 γys o, and On the sides of the channel, C 2 γys o are determined. CVE 372 Hydromechanics 67/68

68 Design of Open Channels Design of Erodible Channels Design Procedure: 2. OPEN CHANNEL FLOW 4. The value of y is determined from the following inequalities and its smaller value is accepted. C 2 sin φ γys ( τ ) = τ. o b = τ permissible p K p 1 2 sin θ 2 C γ ys o τ = 1 p b ( τ ) permissible CVE 372 Hydromechanics 68/68

69 Design of Open Channels Design of Erodible Channels Design Procedure: 5. Using the assumed b/y ratio and the computed value of y the channel capacity (discharge) is determined by the Manning equation Q = A n R 2/3 S o 1/ 2 6. If the computed channel capacity is different from the design discharge, a new value for b/y ratio is assumed and the procedure is repeated until the computed discharge is equal to the given design discharge. 7. Critical flow conditions are checked 8. A freeboard is added on the water depth. CVE 372 Hydromechanics 69/68

70 Design of Open Channels Design of Erodible Channels Note: For cohesive soils, only the tractive force of bottom is critical For noncohesive soils, one must compute the value of K from its equation. On the basis of stability criteria, if: K < 0.78 side shear controls the required depth K > 0.78 bed shear controls the required depth CVE 372 Hydromechanics 70/68

71 Design of Open Channels Design of Erodible Channels Example 18: Design an unlined trapezoidal canal using the tractive force approach to carry Q = 4.6 m3/s. Given are z = 2.0, n = 0.02 and So = The canal material is moderately rounded noncohesive soil that 25% material by weight is larger than 25 mm. CVE 372 Hydromechanics 71/68

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