SGChapt10

SGChapt10 - Ch. 10 Open-Channel Flow Previous internal flow...

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Unformatted text preview: Ch. 10 Open-Channel Flow Previous internal flow analyses have considered only closed conduits where the fluid typically fills the entire conduit and may be either a liquid or a gas. This chapter considers only partially filled channels of liquid flow referred to as open-channel flow. Open-Channel Flow: Flow of a liquid in a conduit with a free surface. Open-channel flow analysis basically results in the balance of gravity and friction forces. One Dimensional Approximation While open-channel flow can, in general, be very complex ( three dimensional and transient), one common approximation in basic analyses is the One-D Approximation: The flow at any local cross section can be treated as uniform and at most varies only in the principal flow direction. This results in the following equations. Conservation of Mass (for ρ = constant) Q = V(x) A(x) = constant Energy Equation 2 2 V1 V + Z1 = 2 + Z 2 + h f 2g 2g | v v X-1 | e-Text Main Menu | Textbook Table of Contents | Study Guide The equation in this form is written between two points ( 1 – 2 ) on the free surface of the flow. Note that along the free surface, the pressure is a constant, is equal to local atmospheric pressure, and does not contribute to the analysis with the energy equation. The friction head loss hf is analogous to the head loss term in duct flow, Ch. VI, and can be represented by where P = wetted perimeter 2 x − x Vavg hf = f 2 1 D h 2g Dh = hydraulic diameter = 4A P Note: One of the most commonly used formulas uses the hydraulic radius: 1 A R h = Dh = 4 P Flow Classification by Depth Variation The most common classification method is by rate of change of free-surface depth. The classes are summarized as 1. Uniform flow (constant depth and slope) 2. Varied flow a. Gradually varied (one-dimensional) b. Rapidly varied (multidimensional) Flow Classification by Froude Number: Surface Wave Speed A second classification method is by the dimensionless Froude number, which is a dimensionless surface wave speed. For a rectangular or very wide channel we have Fr = and V V = 1/2 co (gy) where y is the water depth co = the speed of a surface wave as the wave height approaches zero. There are three flow regimes of incompressible flow. These have analogous flow regimes in compressible flow as shown below: | v v X-2 | e-Text Main Menu | Textbook Table of Contents | Study Guide Incompressible Flow Fr < 1 Fr = 1 Fr > 1 Compressible Flow subcritical flow critical flow supercritical flow Ma < 1 Ma = 1 Ma > 1 subsonic flow sonic flow supersonic flow Hydraulic Jump Analogous to a normal shock in compressible flow, a hydraulic jump provides a mechanism by which an incompressible flow, once having accelerated to the supercritical regime, can return to subcritical flow. This is illustrated by the following figure. Fig. 10.5 Flow under a sluice gate accelerates from subcritical to critical to supercritical and then jumps back to subcritical flow. The critical depth Q yc = 2 b g 1/3 is an important parameter in open- channel flow and is used to determine the local flow regime (Sec. 10.4). Uniform Flow; the Chezy Formula Uniform flow 1. Occurs in long straight runs of constant slope 2. The velocity is constant with V = Vo 3. Slope is constant with So = tan θ | v v X-3 | e-Text Main Menu | Textbook Table of Contents | Study Guide From the energy equation with V1 = V2 = Vo, we have h f = Z1 − Z 2 = S o L Since the flow is fully developed, we can write from Ch. VI 1/2 2 L Vo hf = f Dh 2g 8g 1/2 1/2 Vo = Rh So f and 1/2 8g For fully developed, uniform flow, the quantity is a constant f and can be denoted by C. The equations for velocity and flow rate thus become Vo = C R h So 1/2 1/2 Q = CA R h So 1/2 and 1/2 1/2 The quantity C is called the Chezy coefficient, and varies from 60 ft /s for small 1/2 1/2 rough channels to 160 ft /s for large rough channels (30 to 90 m /s in SI). The Manning Roughness Correlation The friction factor f in the Chezy equations can be obtained from the Moody chart of Ch. VI. However, since most flows can be considered fully rough, it is appropriate to use Eqn 6.64: fully rough flow: 3.7Dh f ≈ 2.0 log ε −2 However, most engineers use a simple correlation by Robert Manning: S.I. Units Vo (m/s ) ≈ B.G. Units Vo (ft/s ) ≈ α [R (m)] nh α 2/3 [R (ft )] nh 2/3 S1/2 o S1/2 o | v v X-4 | e-Text Main Menu | Textbook Table of Contents | Study Guide where n is a roughness parameter given in Table 10.1 and is the same in both systems of units and α is a dimensional constant equal to 1.0 in S.I. units and 1.486 in B.G. units. The volume flow rate is then given by Uniform flow Q = Vo A ≈ α n A R 2/3 S1/2 h o Table 10.1 Experimental Values for Manning’s n Factor | v v X-5 | e-Text Main Menu | Textbook Table of Contents | Study Guide Example 10.1 Given: Rectangular channel, finished concrete, slope = 0.5˚ water depth: y = 4 ft, width: b = 8 ft Find: 3 Volume flow rate (ft /s) 4 ft cross-section 8 ft θ For the given conditions: n = 0.012 A = b y = (8 ft) (4 ft) = 32 ft 2 2 A 32 ft Rh = = = 2 ft P 16 ft So = tan 0.5˚ = 0.0873 P = b + 2 y = 8 + 2 (4) = 16 ft D h = 4 R h = 8 ft Using Manning’s formula in BG units, we obtain for the flow rate Q≈ 1.486 (32 ft 2 )(2 ft )2/3 (0.00873)1.2 ≈ 590 ft 3/s ans. 0.012 Alternative Problem The previous uniform problem can also be formulated where the volume flow rate Q is given and the fluid depth is unknown. For these conditions, the same basic equations are used and the area A and hydraulic radius Rh are expressed in terms of the unknown water depth yn. The solution is then obtained using iterative or systematic trial and error techniques that are available in several math analysis/ math solver packages such as EES (provided with the text) or Mathcad ®. | v v X-6 | e-Text Main Menu | Textbook Table of Contents | Study Guide Uniform Flow in a Partly Full, Circular Pipe Fig. 10.6 shows a partly full, circular pipe with uniform flow. Since frictional resistance increases with wetted perimeter, but volume flow rate increases with cross sectional flow area, the maximum velocity and flow rate occur before the pipe is completely full. For this condition, the geometric properties of the flow are given by the equations below. Fig. 10.6 Uniform Flow in a Partly Full, Circular Channel sin2θ 2 A=R θ− 2 P = 2 Rθ Rh = R sin2θ 1 − 2θ 2 The previous Manning formulas are used to predict Vo and Q for uniform flow when the above expressions are substituted for A, P, and Rh. α R sin2θ Vo ≈ 1 − n 2 2θ 2 /3 1/2 So Q = Vo R 2 sin2θ θ− 2 These equations have respective maxima for Vo and Q given by | v v X-7 | e-Text Main Menu | Textbook Table of Contents | Study Guide α Vmax = 0.718 R 2/3 S1/2 o n Q max = 2.129 α n at θ = 128.73Þ and y = 0.813 D R 8/3 S1/2 o at θ = 151.21Þ and y = 0.938 D Efficient Uniform Flow Channels A common problem in channel flow is that of finding the most efficient lowresistance sections for given conditions. This is typically obtained by maximizing Rh for a given area and flow rate. This is the same as minimizing the wetted perimeter. Note: Minimizing the wetted perimeter for a given flow should minimize the frictional pressure drop per unit length for a given flow. It is shown in the text that for constant value of area A and α = cot θ, the minimum value of wetted perimeter is obtained for [ A = y 2 2 (1 + α 2 ) − α 1/ 2 P = 4 y (1 + α ) 2 1/ 2 − 2α y Rh = 1 y 2 Note: For any trapezoid angle, the most efficient cross section occurs when the hydraulic radius is one-half the depth. For the special case of a rectangle (a = 0, q = 90˚), the most efficient cross section occurs with A = 2y 2 P = 4y Rh = 1 y 2 b = 2y | v v X-8 | e-Text Main Menu | Textbook Table of Contents | Study Guide Best Trapezoid Angle The general equations listed previously are valid for any value of α. For a given, fixed value of area A and depth y the best trapezoid angle is given by α = cot θ = 1 3 θ = 60 o or 1/2 Example 10.3 What are the best dimensions for a rectangular brick channel designed to carry 5 3 m /s of water in uniform flow with So = 0.001? 2 Taking n = 0.015 from Table 10.1, A = 2 y , and Rh = 1/2 y ; Manning’s formula is written as 2/3 1.0 2 1 or 5 m / s = (2 y ) 2 y (0.001)1/ 2 0.015 1.0 2/3 1/2 Q≈ A R h So n 3 This can be solved to obtain y 8/ 3 = 1.882 m 8/3 or y = 1.27 m The corresponding area and width are A = 2 y 2 = 3.21 m2 and b= A = 2.53 m y Note: The text compares these results with those for two other geometries having the same area. | v v X-9 | e-Text Main Menu | Textbook Table of Contents | Study Guide Specific Energy: Critical Depth 2 One useful parameter in channel flow is the specific energy E, where y is the local water depth. V E = y+ 2g Defining a flow per unit channel width as q = Q/b we write q E = y+ 2 g y2 2 Fig. 10.8b is a plot of the water depth y vs. the specific energy E. The water depth for which E is a minimum is referred to as the critical depth yc. Fig. 10.8 Specific Energy Illustration q2 y = yc = g Emin occurs at 1/3 Q2 = 2 b g E min = The value of Emin is given by 1/3 3 y 2c At this value of minimum energy and minimum depth we can write Vc = (g y c ) = Co 1/2 and Fr = 1 | v v X-10 | e-Text Main Menu | Textbook Table of Contents | Study Guide Depending on the value of Emin and V, one of several flow conditions can exist. For a given flow, if E < Emin No solution is possible E = Emin Flow is critical, y = yc, V = Vc E > Emin , V < Vc Flow is subcritical, y > yc ,disturbances can propagate upstream as well as downstream E > Emin , V > Vc Flow is supercritical, y < yc , disturbances can only propagate downstream within a wave angle given by C (g y ) µ = sin o = sin-1 V V 1/ 2 -1 Nonrectangular Channels For flows where the local channel width varies with depth y, critical values can be expressed as bo Q 2 Ac = g 1/3 and Q g Ac Vc = = A c bo 1/2 where bo = channel width at the free surface. These equations must be solved iteratively to determine the critical area Ac and critical velocity Vc. For critical channel flow that is also moving with constant depth (yc), the slope corresponds to a critical slope Sc given by | v v X-11 | e-Text Main Menu | Textbook Table of Contents | Study Guide 2 n gAc Sc = 2 α b o Rh, c α = 1. for S I units and 2.208 for B. G. units and Example 10.5 Given: a 50˚, triangular channel has a 3 flow rate of Q = 16 m /s. Compute: (a) yc, (b) Vc, (c) Sc for n = 0.018 a. For the given geometry, we have P = 2 ( y csc 50˚) A = 2[y (1/2 y cot 50˚)] Rh = A/P = y/2 cos 50˚ bo = 2 ( y cot 50˚) For critical flow, we can write g Ac = bo Q 3 2 or g(yc cot 50Þ = (2 yc cot 50Þ Q ) ) 2 yc = 2.37 m 2 ans. b. With yc, we compute Pc = 6.18 m 2 Ac = 4.70 m bo,c = 3.97 m 3 The critical velocity is now Q 16 m / s Vc = = = 3.41 m / s A 4.70 m ans. c. With n = 0.018, we compute the critical slope as g n2 P 9.81(0.018)2 (6.18) Sc = 2 = = 0.0542 α b o R1/3 1.0 2 (3.97) (0.76)1 /3 h | v v X-12 | e-Text Main Menu | Textbook Table of Contents | Study Guide Frictionless Flow over a Bump Frictionless flow over a bump provides a second interesting analogy, that of compressible gas flow in a nozzle. The flow can either increase or decrease in depth depending on whether the initial flow is subcritical or supercritical. The height of the bump can also change the results of the downstream flow. Fig. 10.9 Frictionless, 2-D flow over a bump Writing the continuity and energy equations for two dimensional, frictionless flow between sections 1 and 2 in Fig. 10.10, we have V1 y1 = V2 y2 2 and 2 V1 V + y1 = 2 + y2 + ∆ h 2g 2g Eliminating V2, we obtain 2 2 2 Vy V y − E2 y + 1 1 = 0 where E2 = 1 + y1 − ∆ h 2g 2g 3 2 2 2 The problem has the following solutions depending on the initial flow condition and the height of the jump: | v v X-13 | e-Text Main Menu | Textbook Table of Contents | Study Guide Key Points: 1. The specific energy E2 is exactly ∆h less than the approach energy E1. 2. Point 2 will lie on the same leg of the curve as point 1. 3. For Fr < 1, subcritical approach The water level will decrease at the bump. Flow at point 2 will be subcritical. 4. For Fr > 1, supercritical approach The water level will increase at the bump. Flow at point 2 will be supercritical. 5. For bump height equal to ∆hmax = E1 - Ec Flow at the crest will be exactly critical (Fr =1). 6. For ∆h > ∆hmax No physically correct, frictionless solutions are possible. Instead, the channel will choke and typically result in a hydraulic jump. Flow under a Sluice Gate A sluice gate is a bottom opening in a wall as shown below in Fig. 10.10a. For free discharge through the gap, the flow smoothly accelerates to critical flow near the gap and the supercritical flow downstream. Fig. 10.10 Flow under a sluice gate This is analogous to the compressible flow through a converging-diverging nozzle. For a free discharge, we can neglect friction. Since this flow has no bump (∆h = 0) and E1 = E2, we can write | v v X-14 | e-Text Main Menu | Textbook Table of Contents | Study Guide 2 V1 2 2 V12 y1 y − + y1 y2 + =0 2g 2g 3 2 This equation has the following possible solutions. Subcritical upstream flow and low to moderate tailwater (downstream water level) One positive, real solution. Supercritical flow at y2 with the same specific energy E2 = E1. Flow rate varies as y2/y1. Maximum flow is obtained for y2/y1 = 2/3. Subcritical upstream flow and high tailwater The sluice gate is drowned or partially drowned (analogous to a choked condition in compressible flow). Energy dissipation will occur downstream in the form of a hydraulic jump and the flow downstream will be subcritical. The Hydraulic Jump The hydraulic jump is an irreversible, frictional dissipation of energy which provides a mechanism for supercritical flow to transition (jump) to subcritical flow analogous to a normal shock in compressible flow. The development of the theory is equivalent to that for a strong fixed wave (Sec 10.1) and is summarized for a hydraulic jump in the following section. | v v X-15 | e-Text Main Menu | Textbook Table of Contents | Study Guide Theory for a Hydraulic Jump If we apply the continuity and momentum equations between points 1 and 2 across a hydraulic jump, we obtain 1 /2 2 y2 = − 1 + (1 + 8 Fr12 ) y1 which can be solved for y2. V2 = We obtain V2 from continuity: V1 y1 y2 The dissipation head loss is obtained from the energy equation as V12 V22 h f = E1 − E2 = + y1 − + y2 2 g 2g or (y − y1 ) =2 3 hf 4 y1 y2 Key points: 1. Since the dissipation loss must be positive, y2 must be > y1. 2. The initial Froude number Fr1 must be > 1 (supercritical flow). 3. The downstream flow must be subcritical and V2 < V1. Example 10.7 3 Water flows in a wide channel at q = 10 m /(s m) and y1 = 1.25 m. If the flow undergoes a hydraulic jump, compute: (a) y2, (b) V2, (c) Fr2, (d) hf, (e) the percentage dissipation, (f) power dissipated/unit width, and (g) temperature rise. | v v X-16 | e-Text Main Menu | Textbook Table of Contents | Study Guide a. The upstream velocity is 3 q 10 m /(s ⋅ m) V1 = = = 8.0 m / s y1 1.25 m Fr1 = The upstream Froude number is 8.0 V1 1/ 2 = 1/ 2 = 2.285 (g y1 ) [9.81 (1.25)] This is a weak jump and y2 is given by 2 y2 2 1/ 2 = − 1 + ( + 8 (2.285) ) = 5.54 1 y1 and y2 = 1 / 2 y1 (5.54) = 3.46 m V2 = b. The downstream velocity is V1 y1 8.0 (1.25) = = 2.89 m / s y2 3.46 c. The downstream Froude number is Fr2 = 2.89 V2 1/ 2 = 1/ 2 = 0.496 9.81 (3.46)] (g y2 ) [ and Fr2 is subcritical as expected. d. The dissipation loss is given by (y − y1 ) =2 3 hf 4 y1 y2 = (3.46 − 1.25)3 = 0.625 m 4 (3.46) (1.25) e. The percentage dissipation is the ratio of hf/E1. 2 2 8.0 V E1 = 1 + y1 = 1.25 + = 4.51 m 2g 2 (9.81) | v v X-17 | e-Text Main Menu | Textbook Table of Contents | Study Guide The percentage loss is thus given by % Loss = hf E1 100 = 0.625 100 = 14% 4.51 f. The power dissipated per unit width is 3 3 Power = ρ Q g hf = 9800 M/m *10 m /(s m) * 0.625 m = 61.3 kw/m g. Using Cp = 4200 J/kg K, the temperature rise is given by & Power dissipated = mCp ∆ T or 61,300 W/m = 10,000 kg/s m * 4200 J/kg K* ∆T ∆T = 0.0015˚K negligible temperature rise | v v X-18 | e-Text Main Menu | Textbook Table of Contents | Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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