HYDRAULICS 14: Open Channel PREPARED BY: ENGR. BILLY I. REJUSO OPEN CHANNEL - It is a conduit in which a liquid flows with a free surface. - An open channel is one which the stream is not completely enclosed by solid boundaries and therefore has a free surface subjected only to atmospheric pressure. - The flow in open channels may either be uniform or non-uniform.
Specific Energy (H) – defined as the energy per unit weight relative to the bottom of the channel. 𝑣2 𝐻= +𝑑 2𝑔 Chezy’s Formula: 1
𝑉 = 𝐶 (𝑅𝑆)2 1
𝑄 = 𝐴𝑉 = 𝐴𝐶 (𝑅𝑆)2 Where: R = Hydraulic Radius = A/P = Cross-section area of the channel/Wetted Perimeter of channel C = Chezy’s coefficient V = Velocity of water (uniform flow) S = Slope of energy gradient m= Bazin Coefficient Re= Reynolds Number ε= Roughness in meter
Kutter’s and Ganguillet Formula: 𝐶=
1 0.00155 23 + 𝑛 + 𝑆
𝑛 0.00155 (23 + ) 𝑆 √𝑅 1.811 0.00281 + 41.65 + 𝑛 𝑆 𝐶= 𝑛 0.00281 1+ (41.65 + ) 𝑆 √𝑅 1+
Manning’s Formula: 1 1 𝑅6 𝑛 1.486 1 𝐶= 𝑅6 𝑛 1 2 1 𝑣 = 𝑅3𝑆2 𝑛 1 2 1 𝑄 = 𝐴 𝑅3𝑆2 𝑛 𝐶=
Bazin’s Formula: 87 𝑚 1+ √𝑅 87 𝐶= 𝑚 0.552 + √𝑅 𝐶=
Powell Equation: 𝐶 𝜀 𝐶 = −42log( + ) 𝑅𝑒 𝑅 Uniform Flow: The simplest of all open channel problem is the uniform flow condition. For the flow to be uniform the following conditions must be satisfied: 1. V1 = V2 2. D1 = D2 3. A1 = A2 4. The stream surface is parallel to the channel bed and the energy grade line is parallel to the stream surface. 5. The slope of the energy grade line S is equal o the slope of the channel bed So Boundary Shear Stress (τo) The average boundary shear stress acting over the wetted surface of the channel is given by: 𝜏𝑜 = 𝛾𝑅𝑆
Sample Problems: 1. Water flows uniformly in a rectangular concrete channel that is 8m wide at a depth of 3m. The channel slope is 0.0025 and using n=0.013 determine the velocity, flow rate and boundary shear stress of the channel. 2. An Earth Canal carries water at a depth of 1.8m. The canal is 6m wide on the bottom and has sides slope 1.5 horizontal to 1 vertical. S= 0.0002. A. Using the value of n=0.025, compute the discharge by Manning’s Formula. B. With the value of discharge obtained from “A, determine the value of “n” in the Kutter’s Formula. Use C=1/n R1/6 by Manning’s Formula. C. Determine the value of “m” in the Bazin Formula. 3. A 600-mm radius sewer pipe is laid on a slope of 0.001 and has a roughness coefficient n=0.012, was found to be 7/8 full. Determine the discharge through the pipe.
HYDRAULICS 15: Most Economic Sections (MES) PREPARED BY: ENGR. BILLY I. REJUSO Normal Depth The normal depth dn is the depth at which uniform flow will occur in an open channel. Normal depth may be determined from Chezy Formula with S=SO. The resulting equation usually requires trial and error solution. Most Economic/Efficient Section of Channels: A section of a channel is said to be most economical when the cost of construction of the channel is minimum. To keep the cost down or minimum, the wetted perimeter, for a given discharge, should be minimum. This condition utilized for determining the dimensions of economical sections of different forms of channels. These are also sections which for a given slope, S, channel cross-sectional area, A, and roughness coefficient, n, the rate of discharge is a maximum. A. Rectangular Section
𝑑 2 𝑏 = 2𝑑 Rectangular channel will be most economic when either: 1. The depth of the flow is half the width 2. The hydraulic radius is half the depth of flow B. Trapezoidal Section 𝑅=
𝑥 = 2𝑦 𝑑 𝑅= 2 Minimum seepage of Trapezoidal Channel: 𝑏 = 4𝑑𝑡𝑎𝑛
ɵ 2
Trapezoidal channel will be most economical when either: 1. One of the sloping sides (wetted length) = half of the top width 2. The hydraulic radius is half the depth of flow C. Triangular Section
𝐴 = 𝑑2 ɵ = 90° D. Circular Section
𝐷 = 0.938𝑑 𝐷 = 0.81𝐷
E. Semi-circular Section
𝑅=
𝑟 2
Sample Problems: 1. A rectangular canal, 6.5m wide and 1.4m deep lined with smooth stone (n=0.013) has a hydraulic slope of 0.001. What savings in earth excavation and lining per meter length of canal could have been effected by using the best proportion of rectangular canal section adhering to the same discharge? 2. A trapezoidal canal has the following flow and channel properties: Discharge=17m3/sec; mean velocity of flow=80cm/sec; roughness coefficient=0.018; side slope=4H:3V. Using the most efficient trapezoidal canal, what slope channel is required? 3. Determine the maximum flow through a 1.2m diameter concrete culvert which is laid on a slope of 0.009. Use n=0.013.
HYDRAULICS 16: Alternate Stages of flow, Non-uniform flow and Hydraulic Jump PREPARED BY: ENGR. BILLY I. REJUSO
Froude Number The ratio of the inertia force to gravity force and is given by expression: 𝑉 𝐹= √𝑔𝐿 For rectangular channel, L =depth of flow d 𝑉 𝐹= √𝑔𝑑 Alternate Stages of Flow For a given total specific energy H, for an open channel flow, there exist two stages or depths of flow that will give the same discharge. These are the upper stage and the lower stage.
Upper Stage - Flow is tranquil - Depth is called subcritical depth - Froude Number, F<1 Lower Stage - Flow is rapid or shooting - Depth is called supercritical depth - Froude Number, F>1
Critical Depth Critical depth is the depth at which for a given total specific energy H, the discharge is maximum, or it is the depth at which for a given discharge Q, the total specific energy is minimum. Critical depth is characterized by: 1. Critical Velocity 2. Critical Slope 3. Froude Number, F=1
𝑄 = 𝐴√2𝑔(𝐻 − 𝑑) Critical depth for rectangular section 3 𝑞2 2 𝑑𝑐 = √ = 𝐻 𝑔 3
Critical depth for any section 𝑄 2 𝐴3 = 𝑔 𝐵 NOTE: A and B are always in terms of d, except that for rectangular canal B is constant.
Where: q= unit discharge (m3/sec per meter width) dc= critical depth Sc= critical slope A= cross-sectional area of flow B= flow width at the top Non-uniform Flow (S≠So) Uniform flow rarely occurs in natural streams because of changes in depth, width, and slope along the channel. The Manning equation for uniform flow can be applied to non-uniform flow with an accuracy dependent on the length of reach taken. 𝐻2 − 𝐻1 𝐿= 𝑆𝑜 − 𝑆 𝑛𝑣𝑚 𝑆 = ( 2 )2 3 𝑅𝑚
Where: Vm= mean velocity between the section 1 and 2 Rm= mean hydraulic radius Hydraulic Jump A hydraulic jump occurs when the upstream flow is supercritical (F>1). According to Chow (1959), a strong jump occurs when F1>9, a steady jump occurs when 4.5
General Equation 𝐴2 ℎ2 − 𝐴1 ℎ1 =
𝑄2 1 1 ( − ) 𝑔 𝐴1 𝐴2
For rectangular canal 𝑞 2 𝑑1 𝑑2 (𝑑1 + 𝑑2 ) = 𝑔 2 𝐹1 − 1 𝐿 = 220𝑑1 𝑡𝑎𝑛ℎ 22 Where: q= vd = Q/b F1= Froude number at section 1
Headlost in the Jump
𝑉12 𝑉22 + 𝑑1 ) − ( + 𝑑2 ) 2𝑔 2𝑔 For rectangular canal, the head lost may be computed by (𝑑2 − 𝑑1 )3 𝐻𝐿 = 4𝑑1 𝑑2 𝐻𝐿 = (
Sample Problems: 1. Water is flowing at a depth of 1.5m in a 3-m wide rectangular channel having n=0.013 and S=0.0009. What is the stage of flow? Determine the critical depth and determine the critical slope. 2. A rectangular concrete channel 4.5m wide carrying water. At an upstream point, the depth of water is 1.5m and at a downstream point 300m away, the depth of flow is 1.17m. If the channel bed slope is 0.0010, find the approximate flow rate. Use n-0.013 3. A parabolic canal has top width of 2m and depth of 2m. Determine the critical depth and critical velocity when the flow is 3m3/sec. 4. A spillway 1.5m deep goes over a dam 30m high. Using a weir factor of 1.92, determine the following: a. The depth of flow at the foot of the spillway and after the jump b. The velocity of flow after the jump c. The energy after the jump d. Head lost in the jump
