FUNDAMENTALSFUNDAMENTALS OFOF FLUIDFLUID MECHANICSMECHANICS ChapterChapter 1010 FlowFlow inin OpenOpen ChannelsChannels
JyhJyh--CherngCherng ShiehShieh Department of Bio-Industrial Mechatronics Engineering National Taiwan University 1 MAINMAIN TOPICSTOPICS
GeneralGeneral CharacteristicsCharacteristics ofof OpenOpen--ChannelChannel FlowFlow SurfaceSurface WavesWaves EnergyEnergy ConsiderationsConsiderations UniformUniform DepthDepth ChannelChannel FlowFlow GraduallyGradually VariesVaries FlowFlow RapidlyRapidly VariesVaries FlowFlow
2 IntroductionIntroduction
Open channel flow involves the flows of a liquid in a channel or conduit that is not completely filled. There exists a free surface between the flowing fluid (usually water) and fluid above it (usually the atmosphere). The main deriving force is the fluid weight-gravity forces the fluid to flow downhill. Under steady, fully developed flow conditions, the component if the weight force in the direction of flow is balanced by the equal and opposite shear force between the fluid and the channel surface.
3 OpenOpen ChannelChannel FlowFlow vs.vs. PipePipe FlowFlow
There can be no pressure force driving the fluid through the channel or conduit. For steady, fully developed channel flow, the pressure distribution within the fluid is merely hydrostatic.
4 ExamplesExamples ofof OpenOpen ChannelChannel FlowFlow
The natural drainage of water through the numerous creek and river systems. The flow of rainwater in the gutters of our houses. The flow in canals, drainage ditches, sewers, and gutters along roads. The flow of small rivulets, and sheets of water across fields or parking lots. The flow in the chutes of water rides.
5 VariablesVariables inin OpenOpen--ChannelChannel FlowFlow
Cross-sectional shape. Bends. Bottom slope variation. Character of its bounding surface. Most open-channel flow results are based on correlation obtained from model and full-scale experiments. Additional information can be gained from various analytical and numerical efforts.
6 GeneralGeneral CharacteristicsCharacteristics ofof OpenOpen--ChannelChannel FlowFlow
7 ClassificationClassification ofof OpenOpen--ChannelChannel FlowFlow
For open-channel flow, the existence of a free surface allows additional types of flow. The extra freedom that allows the fluid to select its free-surface location and configuration allows important phenomena in open- channel flow that cannot occur in pipe flow. The fluid depth, y, varies with time, t, and distance along the channel, x, are used to classify open-channel flow:
8 ClassificationClassification -- TypeType II
Uniform flow (UF): The depth of flow does not vary along the channel (dy/dx=0). Nonuniform flows: ►Rapidly varying flows (RVF): The flow depth changes considerably over a relatively short distance dy/dx~1. ►Gradually varying flows (GVF): The flow depth changes slowly with distance dy/dx <<1.
9 ClassificationClassification -- TypeType IIII
R = ρVR / μ V is the average velocity of the fluid. e h Rh is the hydraulic radius of the channel.
Laminar flow: Re < 500. Transitional flow: Turbulent flow: Re > 12,500.
Most open-channel flows involve water (which has a fairly small viscosity) and have relatively large characteristic lengths, it is uncommon to have laminar open-channel flows.
10 ClassificationClassification -- TypeType IIIIII
Fr = V / gl
Critical Flow: Froude number Fr =1. Subcritical Flow: Froude number Fr <1. Supercritical Flow: Froude number Fr >1.
11 SurfaceSurface WaveWave
12 SurfaceSurface WaveWave
The distinguishing feature of flows involve a free surface (as in open-channel flows) is the opportunity for the free surface to distort into various shapes. The surface of a lake or the ocean is usually distorted into ever- changing patterns associated with surface waves.
13 KindsKinds ofof SurfaceSurface WaveWave
Some of the surface waves are very high, some barely ripple the surface; some waves are very long, some are short; some are breaking wave that form white caps, others are quite smooth. Dsmall amplitude DFinite-sized solitary DContinuous sinusoidal shape
14 smallsmall amplitudeamplitude WaveWave SpeedSpeed 1/51/5
Consider a single elementary wave of small height, by δy, is produced on the surface of a channel by suddenly moving the initially stationary end wall with speed δV.
15 smallsmall amplitudeamplitude WaveWave SpeedSpeed 2/52/5
The water in the channel was stationary at the initial time, t=0. A stationary observer will observe a single wave move down the channel with a wave speed c, with no fluid motion ahead of the wave and a fluid velocity of δV behind the wave. The motion is unsteady. For a observer moving along the channel with speed c, the flow will appear steady. v v To this observer, the fluid velocity will be V = -c i on the observer’s v v right and V = (-c + δ V) i to the left of the observer.
MomentumMomentum EquationEquation ++ ContinuityContinuity EquationEquation
16 smallsmall amplitudeamplitude WaveWave SpeedSpeed 3/53/5
With the assumption of uniform one-dimensional flow, the continuity equation becomes − cyb = (−c + δV)(y + δy)b
(y + δy)δV δV ⇒ c = = y (1) δy δy δy << y Similarly, the momentum equation 1 1 F = γy A = γy2b F = γy A = γ(y + δy)2 b 2 c2 2 2 1 c1 1 2 1 1 F − F = γy2b − γ(y + δy)2 b = ρbcy[](c − δV) − c 2 1 2 2 17 smallsmall amplitudeamplitude WaveWave SpeedSpeed 4/54/5
(δy)2 << yδy δV g ⇒ = (2) δy c
(1)+(2) c = gy (3)
EnergyEnergy EquationEquation ++ ContinuityContinuity EquationEquation
18 smallsmall amplitudeamplitude WaveWave SpeedSpeed 5/55/5
The single wave on the surface is seen by an observer moving with the wave speed, c. Since the pressure is constant at any point on the free surface, the Bernoulli equation for this frictionless flow is V2 VδV + y = cons tan t → + δy = 0 2g g The continuity equation Vy = cons tan t → yδV + Vδy = 0
Combining these two equations and using the fact V=c c = gy 19 finitefinite--sizedsized solitarysolitary WaveWave SpeedSpeed
More advanced analysis and experiments show that the wave speed for finite-sized solitary wave
1/ 2 1/ 2 ⎛ δy ⎞ ⎛ δy ⎞ c = gy⎜1+ ⎟ c = gy⎜1+ ⎟ > gy ⎝ y ⎠ ⎝ y ⎠
The larger the amplitude, the faster the wave travel.
20 ContinuousContinuous sinusoidalsinusoidal shapeshape WaveWave SpeedSpeed 1/21/2
A more general description of wave motion can be obtained by considering continuous (not solitary) wave of sinusoidal shape. DBy combining waves of various wavelengths, λ, and amplitudes, δy. DThe wave speed varies with both the wavelength and fluid depth as
1/ 2 ⎡gλ ⎛ 2πy ⎞⎤ c =& ⎢ tanh⎜ ⎟⎥ (4) ⎣2π ⎝ λ ⎠⎦
21 ContinuousContinuous sinusoidalsinusoidal shapeshape WaveWave SpeedSpeed 2/22/2
1/ 2 ⎡gλ ⎛ 2πy ⎞⎤ c =& ⎢ tanh⎜ ⎟⎥ ⎣2π ⎝ λ ⎠⎦ y ⎛ 2πy ⎞ → ∞ ⇒ tanh⎜ ⎟ = 1 λ ⎝ λ ⎠ gλ ⇒ c = Deep layer 2π
y ⎛ 2πy ⎞ 2πy → 0 ⇒ tanh⎜ ⎟ = λ ⎝ λ ⎠ λ ⇒ c = gy Shallow layer
Wave speed as a function of wavelength.22 FroudeFroude NumberNumber EffectsEffects 1/31/3
Consider an elementary wave travelling on the surface of a fluid. If the fluid layer is stationary, the wave moves to the right with speed c relative to the fluid and stationary observer. When the fluid is flowing to the left with velocity V. DIf V
Fr = V / g = V / c Is the ratio of the fluid l velocity to the wave speed.
23 FroudeFroude NumberNumber EffectsEffects 2/32/3
When a wave is produced on the surface of a moving stream, as happens when a rock is thrown into a river. DIf V=0, the wave speeds equally in all directions. DIf V
24 FroudeFroude NumberNumber EffectsEffects 3/33/3
DIf V=c or Fr=1, the upstream propagating wave remains stationary and the flow is termed critical.
25 EnergyEnergy ConsiderationsConsiderations
26 EnergyEnergy ConsiderationsConsiderations 1/31/3
z1 − z 2 The slope of the channel bottom or bottom S0 = l slope is constant over the segment
Very small for most open-channel flows.
x and y are taken as the distance along the channel bottom and the depth normal to the bottom. 27 EnergyEnergy ConsiderationsConsiderations 2/32/3
With the assumption of a uniform velocity profile across any section of the channel, the one-dimensional energy equation become 2 2 p1 V1 p2 V2 + + z = + + z + h (5) γ 2g 1 γ 2g 2 L
hL is the head loss due to viscous effects between sections (1) and (2). 2 2 V1 V2 (5) y1 + + Sol = y2 + + h L (6) 2g 2g
z1 − z2 = Sol p1 / γ = y1 p2 / γ = y2
28 EnergyEnergy ConsiderationsConsiderations 3/33/3
2 2 V2 − V1 (6) y1 − y2 = + (Sf − So )l (7) 2g
Sf = hL / l
V2 − V2 (7) y − y = 2 1 1 2 2g
For a horizontal channel bottom (S0=0) and negligible head loss (Sf=0)
29 SpecificSpecific EnergyEnergy 1/41/4
Define specific energy, E V2 E = y + (8) 2g
(9) E1 = E2 + (Sf − So )l
(9) E1 + z1 = E2 + z2 The sum of the specific energy and the elevation of the channel bottom remains constant. Head losses are negligible, Sf=0 This a statement of the Bernoulli −Sol = z2 − z1 equation.
30 SpecificSpecific EnergyEnergy 2/42/4
If the cross-sectional shape is a rectangular of width b q2 E = y + (10) 2gy2 Where q is the flowrate per unit width, q=Q/b=Vyb/b=Vy
For a given channel b= constant q = constant E = E (y) Specific energy diagram
31 SpecificSpecific EnergyEnergy 3/43/4
q2 E = y + (10) 2gy2 For a given q and E, equation (10) is a
cubic equation with three solutions , ysup, ysub, and yneg.
If E >Emin, two solutions are positive and yneg is negative (has no physical meaning and can be ignored). These two depths are term alternative depths.
32 SpecificSpecific EnergyEnergy 4/44/4
Approach y=E Very deep and very slowly ysup < ysub
Vsup > Vsub
E > Emin Two possible depths of flow, one subcritical and the other supercritical
Approach y=0 Very shallow and very high speed 33 DetermineDetermine EEminmin
To determine the value of Emin 1/ 3 dE dE q2 ⎛ q2 ⎞ = 0 ⇒ = 1− = 0 ⇒ y = ⎜ ⎟ (11) 3 c ⎜ ⎟ dy dy gy ⎝ g ⎠
3yc q Sub. (11) into (10) Emin = ⇒ Vc = = gyc ⇒ Frc = 1 2 yc
1. The critical conditions (Fr=1) occur at the location of Emin. 2. Flows for the upper part of the specific energy diagram are subcritical (Fr<1) 3. Flows for the lower part of the specific energy diagram are supercritical (Fr>1)
34 ExampleExample 10.110.1 SpecificSpecific EnergyEnergy diagramdiagram -- QualitativeQualitative
Water flows under the sluice gate in a constant width rectangular channel as shown in Fig. E10.1a. Describe this flow in terms of the specific energy diagram. Assume inviscid flow.
35 ExampleExample 10.110.1 SolutionSolution1/21/2
Inviscid flow Sf=0 Channel bottom is horizontal z1=z2 (or So=0)
E1=E2 q1=q2 The specific energy diagram for this flow is as shown in Fig. E10.1b.
The flowrate can remain the same for this channel even if the
upstream depth is increased. This is indicated by depths y1’ and y2’ in Fig E10.1c. To remain same flowrate, the distance between the bottom of the gate and the channel bottom must be decreased to give a smaller
flow area (y2’ < y2 ), and the upstream depth must be increased to give a bigger head (y > y ). give a bigger head (y 1’ > y1 ). 36 ExampleExample 10.110.1 SolutionSolution2/22/2
On the other hand, if the gate remains fixed so that the downstream
depth remain fixed (y2’’ = y2 ), the flowrate will increase as the upstream depth increases to y 1’’ > y1.
q”>q0
37 ExampleExample 10.210.2 SpecificSpecific EnergyEnergy diagramdiagram –– QuantitativeQuantitative
Water flows up a 0.5-ft-tall ramp in a constant width rectangular channel at a rate of q = 5.75 ft2/s as shown in Fig. E10.2a. (For now disregard the “bump”) If the upstream depth is 2.3 ft, determine the
elevation of the water surface downstream of the ramp, y2 + z2. Neglect viscous effects.
38 ExampleExample 10.210.2 SolutionSolution1/41/4
With S0 l =z1-z2 and hL=0, conservation of energy requires that p V2 p V2 1 + 1 + z = 2 + 2 + z γ 2g 1 γ 2g 2 2 V2 ⇒ 1.90 = y + (10.2-1) 2 64.4 3 2 The continuity equation y2 −1.90y2 + 0.513 = 0 y V = y V 2 2 1 1 y2 = 1.72ft 0.638ft 2 (10.2-2) ⇒ y2V2 = 5.75ft /s y2 = −0.466ft
39 ExampleExample 10.210.2 SolutionSolution2/42/4
The corresponding elevations of the free surface are either
y2 + z2 = 1.72ft + 0.50ft = 2.22ft y2 + z2 = 0.638ft + 0.50ft = 1.14ft
Which of these flows is to be expected? This can be answered by use of the specific energy diagram obtained from Eq.(10), which for this problem is 0.513 E = y + y2 The diagram is shown in Fig.E10.2(b).
40 ExampleExample 10.210.2 SolutionSolution3/43/4
41 ExampleExample 10.210.2 SolutionSolution4/44/4
The upstream condition corresponds to subcritical flow; the downstream condition is either subcritical or supercritical, corresponding to points 2 or 2’.
Note that since E1=E2+(z2-z1)=E2+0.5 ft, it follows that the downstream conditions are located to 0.5 ft to the left of the upstream conditions on the diagram.
…….... TheThe surfacesurface elevationelevation isis
y2 + z2 = 1.72ft + 0.50ft = 2.22ft
42 ChannelChannel DepthDepth VariationsVariations 1/31/3
Consider gradually varying flows. For such flows, dy/dx<<1, and it is reasonable to impose the one- dimensional velocity assumption. At an section the total head V2 H = + y + z 2g
and the energy equation H1 = H2 + hL The slop of the energy line 2 dH dh dz dH d ⎛ V ⎞ V dV dy dz L = ⎜ + y + z⎟ = + + = = Sf = So ⎜ ⎟ dx dx dx dx dx ⎝ 2g ⎠ g dx dx dx
43 ChannelChannel DepthDepth VariationsVariations 2/32/3
dh V dV dy L = + + S dx g dx dx 0 V dV dy + = S − S (12) g dx dx f o For a given flowrate per unit width, q, in a rectangular channel of constant width b, we have
q dV q dy V dy V = ⇒ = − = − y dx y2 dx y dx
V dV V2 dy dy = − = −F2 (13) (12) g dx gy dx r dx
44 ChannelChannel DepthDepth VariationsVariations 3/33/3
dy Sf − So Sub. (13) into (12) = (14) Sub. (13) into (12) 2 dx (1− Fr )
DependsDepends onon thethe locallocal slopeslope ofof thethe channelchannel bottom,bottom, thethe slopeslope ofof thethe energyenergy line,line, andand thethe FroudeFroude number.number.
45 UniformUniform DepthDepth ChannelChannel FlowFlow
46 UniformUniform DepthDepth ChannelChannel FlowFlow 1/31/3
Many channels are designed to carry fluid at a uniform depth all along their length. DIrrigation canals. DNature channels such as rivers and creeks. Uniform depth flow (dy/dx=0) can be accomplished by adjusting the
bottom slope, S0, so that it precisely equal the slope of the energy line, Sf. A balance between the potential energy lost by the fluid as it coasts downhill and the energy that is dissipated by viscous effects (head loss) associated with shear stress throughout the fluid.
47 UniformUniform DepthDepth ChannelChannel FlowFlow 2/32/3
Uniform flow in an open channel.
48 UniformUniform DepthDepth ChannelChannel FlowFlow 3/33/3
Typical velocity and shear stress distributions in an open channel: (a) velocity distribution throughout the cross section. (b) shear stress distribution on the wetted perimeter. 49 TheThe ChezyChezy && ManningManning EquationEquation 1/61/6
Control volume for uniform flow in an open channel.
50 TheThe ChezyChezy && ManningManning EquationEquation 2/62/6
Under the assumption of steady uniform flow, the x component of the momentum equation
∑Fx = ρQ(V2 − V1) = 0
F1 − F2 − τw Pl + Wsin θ = 0 (15)
where F1 and F2 are the hydrostatic pressure forces across either end of the control volume. P is wetted perimeter.
51 TheThe ChezyChezy && ManningManning EquationEquation 3/63/6
y1=y2 F1=F2
(15) −τwPl + Wsinθ = 0 Wsinθ A τw = W = γAl R h = Pl P
γAlSo τw = = rR hSo (16) Pl V2 Wall shear stress is proportional to the dynamic pressure τ ∝ ρ w 2 V2 (Chapter 8) τ = Kρ w 2 K is a constant dependent upon the roughness of the pipe
52 TheThe ChezyChezy && ManningManning EquationEquation 4/64/6
V2 (16) Kρ = rR S V = C R S (17) 2 h o h o C is termed the Chezy coefficient Chezy equation
Was developed in 1768 by A. Chezy (1718-1798), a French engineer who designed a canal for the Paris water supply.
1/ 2 (17) V ∝ So Reasonable
2 / 3 V ∝ R h V ∝ R h
ManningManning EquationEquation
53 TheThe ChezyChezy && ManningManning EquationEquation 5/65/6
In 1889, R. manning (1816-1897), an Irish engineer, developed the following somewhat modified equation for open-channel flow to
more accurately describe the Rh dependence:
R 2 / 3S1/ 2 V = h o (18) Manning equation n n is the Manning resistance coefficient. Its value is dependent on the surface material of the channel’s wetted perimeter and is obtained from experiments. It has the units of s/m1/3 or s./ft1/3
54 TheThe ChezyChezy && ManningManning EquationEquation 6/66/6
κ (18) V = R 2 / 3S1/ 2 (19) Where κ=1 if SI units are n h o used, κ=1.49 if BG units κ Q = AR 2 / 3S1/ 2 (20) are used. n h o The best hydraulic cross section is defined as the section of minimum
area for a given flowrate Q, slope, So, and the roughness coefficient, n.
2 / 3 3/5 κ ⎛ A ⎞ κ A5/ 3S1/ 2 ⎡ nQ ⎤ Q = A⎜ ⎟ S1/ 2 = o ⇒ A = ⎢ ⎥ P2 /5 n P o n 2 / 3 1/ 2 ⎝ ⎠ P ⎣⎢ kSo ⎦⎥ A R = constant h P A channel with minimum A is one with a minimum P.
55 ValueValue ofof thethe ManningManning Coefficient,Coefficient, nn
56 UniformUniform DepthDepth ExamplesExamples
57 ExampleExample 10.310.3 UniformUniform Flow,Flow, DetermineDetermine FlowFlow RateRate
Water flows in the canal of trapezoidal cross section shown in Fig. E10.3a. The bottom drops 1.4 ft per 1000 ft of length. Determine the flowrate if the canal is lined with new smooth concrete. Determine the Froude number for this flow.
58 ExampleExample 10.310.3 SolutionSolution1/21/2
κ (20) Q = AR 2 / 3S1/ 2 κ=1.49 if BG units are used. n h o ⎛ 5 ⎞ A = 12ft(5ft) + 5ft⎜ ft⎟ = 89.8ft 2 ⎝ tan 40° ⎠ P = 12ft + 2(5/sin 40°ft) = 27.6ft From Table 10.1, n=0.012 R h = A / P = 3.25ft 1.49 10.98 Q = (89.8ft 2 )(3.25ft)2 / 3(0.0014)1/ 2 = = 915cfs n n V V = Q / A = 10.2ft / s Fr = = ... = 0.804 gy
59 ExampleExample 10.310.3 SolutionSolution2/22/2
60 ExampleExample 10.410.4 UniformUniform Flow,Flow, DetermineDetermine FlowFlow DepthDepth
Water flows in the channel shown in Fig. E10.3 at a rate o Q = 10.0 m3/s. If the canal lining is weedy, determine the depth of the flow.
61 ExampleExample 10.410.4 SolutionSolution
A = 1.19y2 + 3.66y ⎛ y ⎞ P = 3.66 + 2⎜ ⎟ = 3.11y + 3.66 ⎝ sin 40° ⎠ A 1.19y2 + 3.66y R = = h P 3.11y + 3.66 From Table 10.1, n=0.030
κ 1.0 Q = 10 = AR 2 / 3S1/ 2 = ... n h o 0.03 (1.19y2 + 3.66y)5 − 515(3.11y + 3.66)2 = 0
y=1.50y=1.50 mm
62 ExampleExample 10.510.5 UniformUniform Flow,Flow, MaximumMaximum FlowFlow RateRate
Water flows in a round pipe of diameter D at a depth of 0 ≤ y ≤ D, as shown in Fig. E10.5a. The pipe is laid on a constant slope of S0, and the Manning coefficient is n. At what depth does the maximum flowrate occur? Show that for certain flowrate there are two depths possible with the same flowrate. Explain this behavior.
63 ExampleExample 10.510.5 SolutionSolution1/21/2
κ (20) Q = AR 2 / 3S1/ 2 κ=1.49 if BG units are used. n h o D2 A = (θ − sin θ) 8 Dθ A D(θ − sin θ) P = R = = 2 h P 4θ κ D8/ 3 ⎡(θ − sin θ)5/ 3 ⎤ Q = S1/ 2 o 2 / 3 ⎢ 2 / 3 ⎥ n 8(4) ⎣⎢ θ ⎦⎥ This can be written in terms of the flow depth by using D y = [1 − cos(θ/ 2)] 2 64 ExampleExample 10.510.5 SolutionSolution2/22/2
A graph of flowrate versus flow depth, Q = Q(y), has the characteristic indicated in Fig. E10.5(b).
The maximum flowrate occurs when y=0.938D, or θ=303º
Q = Qmax when y = 0.938D
65 ExampleExample 10.610.6 UniformUniform Flow,Flow, EffectEffect ofof BottomBottom SlopeSlope
Water flows in a rectangular channel of width b = 10 m that has a Manning coefficient of n = 0.025. Plot a graph of flowrate, Q, as a
function of slope S0, indicating lines of constant depth and lines of constant Froude number.
66 ExampleExample 10.610.6 SolutionSolution1/21/2
A by A = by = 10y R = = h P (b + 2y) 2 / 3 κ 2 / 3 1/ 2 1.0 ⎛ 10y ⎞ 1/ 2 (19) V = R h So = ⎜ ⎟ So (10.6-1) n 0.025 ⎝ 10 + 2y ⎠ 2 / 3 1/ 2 1.0 ⎛ 10y ⎞ 1/ 2 (gy) Fr = ⎜ ⎟ So 0.025 ⎝ 10 + 2y ⎠ 4 / 3 2 ⎛ 5 + y ⎞ ⇒ So = 0.00613Fr y⎜ ⎟ (10.6-2) ⎝ 5y ⎠
67 ExampleExample 10.610.6 SolutionSolution2/22/2
For given value of Fr, we pick various value of y, determine the
corresponding value of So from Eq (10.6-2), and then calculate Q=VA, with V from either Eq (10.6-1) or V=(gy)1/2Fr.
TheThe resultsresults areare indicatedindicated inin Fig.Fig. E10.6E10.6
68 ExampleExample 10.710.7 UniformUniform Flow,Flow, VariableVariable RoughnessRoughness
Water flows along the drainage canal having the properties shown in
Fig. E10.7a. If the bottom slope is S0 = 1 ft/500 ft=0.002, estimate the flowrate when the depth is y = 0.8 ft + 0.6 ft = 1.4 ft.
69 ExampleExample 10.710.7 SolutionSolution
Q = Q1 + Q2 + Q3
1.49 2 / 3 1/ 2 Qi = AiR h So ni i
Q = ... = 16.8ft3 /s
70 ExampleExample 10.810.8 UniformUniform Flow,Flow, BestBest HydraulicHydraulic CrossCross SectionSection
Water flows uniformly in a rectangular channel of width b and depth y. Determine the aspect ratio, b/y, for the best hydraulic cross section.
71 ExampleExample 10.810.8 SolutionSolution1/31/3
κ (20) Q = AR 2 / 3S1/ 2 κ=1.49 if BG units are used. n h o A = by P = b + 2y A by A Ay R h = = = = P (b + 2y) (b + 2y) (2y2 + A)
2 / 3 κ ⎛ Ay ⎞ Q = A⎜ ⎟ S1/ 2 ⎜ 2 ⎟ o n ⎝ (2y + A) ⎠ 3/ 2 ⎛ nQ ⎞ A5/ 2y = K(2y2 + A) K = ⎜ ⎟ ⎜ 1/ 2 ⎟ ⎝ κSo ⎠
constant 72 ExampleExample 10.810.8 SolutionSolution2/32/3
The best hydraulic section is the one that gives the minimum A for all y. That is, dA/dy = 0.
dA 5 3/ 2 dA 5/ 2 ⎛ dA ⎞ = 0 ⇒ A y + A = K⎜4y + ⎟ dy 2 dy ⎝ dy ⎠ ⇒ A5/ 2 = 4ky
2y2 = by The rectangular with the best hydraulic cross section twice as wide as it is deep, or b / y = 2 73 ExampleExample 10.810.8 SolutionSolution3/33/3
TheThe bestbest hydraulichydraulic crosscross sectionsection forfor otherother shapesshapes
74 GraduallyGradually VariedVaried FlowFlow dy << 1 dx
75 GraduallyGradually VariedVaried FlowFlow 1/21/2
Open channel flows are classified as uniform depth, gradually varying or rapidly varying. If the channel bottom slope is equal to the slope of the energy line,
So=Sf, the flow depth is constant, dy/dx=0. DThe loss in potential energy of the fluid as it flows downhill is exactly balanced by the dissipation of energy through viscous effects. If the bottom slope and the energy line slope are not equal, the flow depth will vary along the channel.
76 GraduallyGradually VariedVaried FlowFlow 2/22/2
dy Sf − So = (14) 2 dx (1− Fr )
TheThe signsign ofof dy/dxdy/dx,, thatthat is,is, whetherwhether thethe flowflow depthdepth increaseincrease oror decreasedecrease withwith distancedistance alongalong thethe channelchannel 2 dependdepend onon SSf --SSo adad 11--FrFr
77 ClassificationClassification ofof SurfaceSurface ShapesShapes 1/31/3
The character of a gradually varying flow is often classified in terms
of the actual channel slope, So, compared with the slope required to produce uniform critical flow, Soc. The character of a gradually varying flow depends on whether the
fluid depth is less than or greater than the uniform normal depth, yn.
1212 possiblepossible surfacesurface configurationsconfigurations
78 ClassificationClassification ofof SurfaceSurface ShapesShapes 2/32/3
Fr<1 : y>yc Fr>1 : y 79 ExamplesExamples ofof GraduallyGradually VariesVaries FlowsFlows 1/5 Backwater curve Drop-down profile Typical surface configurations for nonuniform depth flow with a mild slope. S0 < S0c. 80 ExamplesExamples ofof GraduallyGradually VariesVaries FlowsFlows 2/5 Typical surface configurations for nonuniform depth flow with a critical slope. S0 = S0c. 81 ExamplesExamples ofof GraduallyGradually VariesVaries FlowsFlows 3/5 Typical surface configurations for nonuniform depth flow with a steep slope. S0 > S0c. 82 ExamplesExamples ofof GraduallyGradually VariesVaries FlowsFlows 4/5 Typical surface configurations for nonuniform depth flow with a horizontal slope. S0 =0. 83 ExamplesExamples ofof GraduallyGradually VariesVaries FlowsFlows 5/5 Typical surface configurations for nonuniform depth flow with a adverse slope. S0 <0. 84 ClassificationClassification ofof SurfaceSurface ShapesShapes 3/33/3 The free surface is relatively free to conform to the shape that satisfies the governing mass, momentum, and energy equations. The actual shape of the surface is often very important in the design of open-channel devices or in the prediction of flood levels in natural channels. ☯☯TheThe surfacesurface shape,shape, y=y=y(xy(x),), cancan bebe calculatedcalculated byby solvingsolving thethe governinggoverning differentialdifferential equationequation obtainedobtained fromfrom aa combinationcombination ofof thethe ManningManning equationequation (20)(20) andand thethe energyenergy equationequation (14).(14). NumericalNumerical techniquestechniques havehave beenbeen developeddeveloped andand usedused toto predictpredict openopen--channelchannel surfacesurface shapes.shapes. 85 RapidlyRapidly VariedVaried FlowFlow dy ~ 1 dx 86 RapidlyRapidly VariedVaried FlowFlow Rapidly varied flow: flow depth changes occur over a relatively short distance. DQuite complex and difficult to analyze in a precise fashion. DMany approximate results can be obtained by using a simple one-dimensional model along with appropriate experimentally determined coefficients when necessary. 87 OccurrenceOccurrence ofof RapidlyRapidly VariedVaried FlowFlow 1/21/2 Flow depth changes significantly un a short distance: The flow changes from a relatively shallow, high speed condition into a relatively deep, low speed condition within a horizontal distance of just a few channel depths. Hydraulic Jump 88 OccurrenceOccurrence ofof RapidlyRapidly VariedVaried FlowFlow 2/22/2 Sudden change in the channel geometry such as the flow in an expansion or contraction section of a channel. Rapidly varied flow may occur in a channel transition section. 89 ExampleExample ofof RapidlyRapidly VariedVaried FlowFlow 1/21/2 The scouring of a river bottom in the neighborhood of a bridge pier. Responsible for the erosion near the foot of the bridge pier. The complex three-dimensional flow structure around a bridge pier. 90 ExampleExample ofof RapidlyRapidly VariedVaried FlowFlow 2/22/2 Many open-channel flowflow--measuringmeasuring devicesdevices are based on principles associated with rapidly varied flows. DDBroadBroad--crestedcrested weirs.weirs. DDSharpSharp--crestedcrested weirs.weirs. DDCriticalCritical flowflow flumes.flumes. DDSluiceSluice gates.gates. 91 HydraulicHydraulic JumpJump 92 TheThe HydraulicHydraulic JumpJump 1/61/6 Under certain conditions it is possible that the fluid depth will change very rapidly over a short length of the channel without any change in the channel configuration. DSuch changes in depth can be approximated as a discontinuity in the free surface elevation (dy/dx=∞). DThis near discontinuity is called a hydraulichydraulic jumpjump. . 93 TheThe HydraulicHydraulic JumpJump 2/62/6 A simplest type of hydraulic jump in a horizontal, rectangular channel. Assume that the flow at sections (1) and (2) is nearly uniform, steady, and one-dimensional. 94 TheThe HydraulicHydraulic JumpJump 3/63/6 The x component of the momentum equation F1 − F2 = ρQ(V2 − V) = ρV1y1b(V2 − V1) 2 F1 = pc1A1 = γy1 b / 2 y2 y2 V y 1 − 2 = 1 1 (V − V) (21) 2 2 2 g 2 F2 = pc2A2 = γy2b / 2 The conservation of mass equation y1bV1 = y2bV2 = Q (22) 2 2 V1 V2 The energy equation y1 + = y2 + + hL (23) 2g 2g The head loss is due to the violent turbulent mixing and dissipation. 95 TheThe HydraulicHydraulic JumpJump 4/64/6 (21)+(22)+(23) NonlinearNonlinear equationsequations One solution is y1=y2, V1=V2, hL=0 OtherOther solutions?solutions? 2 2 2 y1 y2 V1y1 ⎛ V1y1 ⎞ V1 y1 (21)+(22) − = ⎜ − V1 ⎟ = (y1 − y2 ) 2 2 g ⎝ y2 ⎠ gy2 2 ⎛ y ⎞ ⎛ y ⎞ V1 ⎜ 2 ⎟ + ⎜ 2 ⎟ − 2F2 = 0 Fr1 = ⎜ ⎟ ⎜ ⎟ r1 gy1 ⎝ y1 ⎠ ⎝ y1 ⎠ y 1 y2 1 ⎛ 2 ⎞ 2 = ⎛−1+ 1+ 8F2 ⎞ (24) SolutionsSolutions = ⎜−1± 1+ 8Fr1 ⎟ ⎜ r1 ⎟ y1 2 ⎝ ⎠ y1 2 ⎝ ⎠ 96 TheThe HydraulicHydraulic JumpJump 5/65/6 2 ⎡ 2 ⎤ hL y2 Fr1 ⎛ y1 ⎞ (23) = 1− + ⎢1− ⎜ ⎟ ⎥ (25) (23) y y 2 ⎢ ⎜ y ⎟ ⎥ 1 1 ⎣ ⎝ 2 ⎠ ⎦ (24)+(25) Depth ratio and dimensionless head loss across a hydraulic jump as a function of upstream Froude number. 97 TheThe HydraulicHydraulic JumpJump 6/66/6 The head loss is negative if Fr1<1. Violate the second law of thermodynamics NotNot possiblepossible toto produceproduce aa hydraulichydraulic jumpjump withwith FrFr1<1.<1. 98 ClassificationClassification ofof HydraulicHydraulic JumpJump 1/21/2 The actual structure of a hydraulic jump is a complex function of Fr1, even though the depth ratio and head loss are given quite accurately by a simple one-dimensional flow analysis. A detailed investigation of the flow indicates that there are essentially five type of surface and jump conditions. 99 ClassificationClassification ofof HydraulicHydraulic JumpJump 2/22/2 100 HydraulicHydraulic JumpJump VariationsVariations 1/21/2 Hydraulic jumps can occur in a variety o channel flow configurations, not justjust inin horizhorizontal, rectangular channels as discussed above. Other common types of hydraulic jumps include those that occur in sloping channels and the submerged hydraulic jumps that can occur just downstream of a sluice gate. 101 HydraulicHydraulic JumpJump VariationsVariations 2/22/2 Hydraulic jump variations: (a) jump caused by a change in channel slope, (b) submerged jump 102 ExampleExample 10.910.9 HydraulicHydraulic JumpJump Water on the horizontal apron of the 100-ft-wide spillway shown in Fig. E10.9a has a depth o 0.60 ft and a velocity of 18 ft/s. Determine the depth, y2, after the jump, the Froude numbers before and after the jump, Fr1 and Fr2, and the power dissipated, Pd, with the jump. 103 ExampleExample 10.910.9 SolutionSolution1/31/3 Conditions across the jump are determined by the upstream Froude number V1 18ft /s Fr1 = = = 4.10 2 gy1 (32.2ft /s )(0.60ft) y2 1 ⎛ 2 ⎞ (24) = ⎜−1+ 1+ 8Fr1 ⎟ = ... = 5.32 y1 2 ⎝ ⎠ y2 = 5.32(0.60ft) = 3.19ft Since Q1=Q2, or V2=(y1V1)/y2=3.39ft/s 104 ExampleExample 10.910.9 SolutionSolution2/32/3 V2 3.39ft /s Fr2 = = = 0.334 2 gy1 (32.2ft /s )(3.19ft) The poser dissipated, Pd, by viscous effects within the jump can be determined from the head loss Pd = γQhL = γby1V1hL ⎛ V2 ⎞ ⎛ V2 ⎞ (23) h = ⎜ y + 1 ⎟ − ⎜ y + 2 ⎟ = ... = 2.26ft L ⎜ 1 ⎟ ⎜ 2 ⎟ ⎝ 2g ⎠ ⎝ 2g ⎠ 5 Pd = γQhL = γby1V1hL = ...1.52 ×10 ft ⋅ lb /s = 277hp 105 ExampleExample 10.910.9 SolutionSolution3/33/3 Q q = q = q = = y V = 10.8ft2 /s 1 2 b 1 1 q2 1.81 E = y + = y + 2gy2 y2 Various upstream depth 106 WeirsWeirs andand GateGate 107 WeirWeir A weir is an obstruction on a channel bottom over which the fluid must flow. Weir provides a convenient method of determining the flowrate in an open channel in terms of a single depth measurement. 108 SharpSharp--CrestedCrested WeirWeir 1/41/4 A sharp-Crested weir is essentially a vertical-edged flat plate placed across the channel. The fluid must flow across the sharp edge and drop into the pool downstream of the weir plate. 109 SharpSharp--CrestedCrested WeirWeir -- GeometryGeometry 2/42/4 Sharp-crested weir plate geometry: (a) rectangular, (b) triangular, (c) trapezoidal. 110 SharpSharp--CrestedCrested WeirWeir –– FlowrateFlowrate 3/43/4 AssumeAssume thatthat thethe velocityvelocity profilprofilee upstreamupstream ofof thethe weirweir plateplate isis uniformuniform andand thatthat ththee pressurepressure withinwithin thethe nappenappe isis atmosphere.atmosphere. AssumeAssume thatthat thethe fluidfluid flowsflows horizontallyhorizontally overover thethe weirweir plateplate withwith aa nonuniformnonuniform velocityvelocity profile.profile. 111 SharpSharp--CrestedCrested WeirWeir –– FlowrateFlowrate 4/44/4 With PB=0, the Bernoulli equation for flow along the arbitrary streamline A-B indicated can be written as p V2 u2 A + 1 + z = (H + P − h) + 2 (26) γ 2g A w 2g Since the total head for any particle along the vertical section (1) is the same p V2 V2 z + A + 1 = H + p + 1 A γ 2g w 2g ⎛ V2 ⎞ h=H (26) u = 2g⎜h + 1 ⎟ (27) 2 ⎜ ⎟ Q = u2dA = u2ldh ⎝ 2g ⎠ ∫ ∫h=0 112 RectangularRectangular WeirWeir –– FlowrateFlowrate 1/21/2 ForFor aa rectangularrectangular weir,weir, ll=b=b 1/ 2 3/ 2 3/ 2 H ⎛ 2 ⎞ ⎧⎛ 2 ⎞ ⎛ 2 ⎞ ⎫ V1 2 ⎪ V1 V1 ⎪ Q = 2gb ⎜h + ⎟ dh = 2gb⎨⎜ H + ⎟ − ⎜ ⎟ ⎬ (28) ∫0 ⎜ 2g ⎟ 3 ⎜ 2g ⎟ ⎜ 2g ⎟ ⎝ ⎠ ⎩⎪⎝ ⎠ ⎝ ⎠ ⎭⎪ 3/ 2 Q = 2 2gbH (29) (28) 3 V2 1 << H 2g Because of the numerous approximations made to obtain Eq. (29) 2 3/ 2 Q = C 2gbH (30) wr 3 (30) 113 RectangularRectangular WeirWeir –– FlowrateFlowrate 2/22/2 Cwr is the rectangular weir coefficient. Cwr is function of Reynolds number (viscous effects), Weber number (surface tension effects), H/Pw (geometry effects). In most practical situations, the Reynolds and Weber number effects are negligible, and the following correction can be used. ⎛ H ⎞ (31) Cwr = 0.611+ 0.075⎜ ⎟ ⎝ Pw ⎠ 114 TriangularTriangular WeirWeir –– FlowrateFlowrate 1/21/2 = 2(H − h) tan θ ForFor aa triangulartriangular weirweir l 2 8 θ Q = tan 2gH5/ 2 15 2 V2 1 << H 2g An experimentally determined triangular weir coefficient, Cwt, is used to account for the real world effects neglected in the analysis so that 8 θ Q = C tan 2gH5/ 2 (32) wt 15 2 115 TriangularTriangular WeirWeir –– FlowrateFlowrate 2/22/2 Weir coefficient for triangular sharp-crested weirs 116 AboutAbout NappeNappe Flowrate over a weir depends on whether the napple is free or submerged. Flowrate will be different for these situations than that give by Eq. (30) and (32). Flow conditions over a weir without a free nappe: (a) plunging nappe, (b) submerged nappe. 117 BroadBroad--CrestedCrested WeirWeir 1/31/3 A broad-crested weir is a structure in an open channel that has a horizontal crest above which the fluid pressure may be considered hydrostatic. 118 BroadBroad--CrestedCrested WeirWeir 2/32/3 Generally, these weirs are restricted to the range 0.08 < H/Lw < 0.50. For long weir block (H/Lw < 0.08), head losses across the weir cannot be neglected. For short weir block (H/Lw > 0.50), the streamlines of the flow over the weir are not horizontal. Apply the Bernoulli equation V2 V2 H + P + 1 = y + P + c w 2g c w 2g If the upstream velocity head is negligible V2 − V2 V2 H − y = c 1 = c c 2g 2g 119 BroadBroad--CrestedCrested WeirWeir 3/33/3 Since V2 = Vc = gyc y 2 H − y = c y = H c 2 c 3 3/ 2 3/ 2 ⎛ 2 ⎞ 3/ 2 The flowrate is Q = by2V2 = bycVc = b gyc = b g⎜ ⎟ H ⎝ 3 ⎠ Again an empirical broad-crested weir coefficient, Cwb, is used to account for the real world effects neglected in the analysis so that 3/ 2 ⎛ 2 ⎞ 3/ 2 0.65 C = Q = Cwbb g⎜ ⎟ H (33) wb 1/ 2 (34) ⎝ 3 ⎠ ⎛ H ⎞ ⎜1+ ⎟ ⎝ Pw ⎠ 120 ExampleExample 10.1010.10 SharpSharp--CrestedCrested andand broadbroad-- CrestedCrested WeirsWeirs Water flows in a rectangular channel of width b = 2 m with flowrate 3 3 between Qmin = 0.02 m /s and Qmax = 0.60 m /s. This flowrate is to be measured by using (a) a rectangular sharp-crested weir, (b) a triangular sharp-crested weir with θ=90º, or (c) a broad-crested weir. In all cases the bottom of the flow area over the weir is a distance Pw = 1 m above the channel bottom. Plot a graph of Q= Q(H) for each weir and comment on which weir would be best for this application. 121 ExampleExample 10.1010.10 SolutionSolution1/31/3 For the rectangular weir with Pw=1. ⎛ H ⎞ (30)+(31) Q = C 2 2gbH 3/ 2 = ⎜0.611 + 0.075 ⎟ 2 2gbH 3/ 2 wr 3 ⎜ ⎟ 3 ⎝ Pw ⎠ Q = 5.91(0.611 + 0.075H)H 3/ 2 For the triangular weir 8 θ 5/ 2 5/ 2 (32) Q = C tan 2gH = 2.36C H wt 15 2 wt 122 ExampleExample 10.1010.10 SolutionSolution2/32/3 For the broad-crested weir 3/ 2 3/ 2 ⎛ 2 ⎞ 0.65 ⎛ 2 ⎞ Q = C b g H3/ 2 = b g H3/ 2 (33)+(34) wb ⎜ ⎟ 1/ 2 ⎜ ⎟ ⎝ 3 ⎠ ⎛ H ⎞ ⎝ 3 ⎠ ⎜1+ ⎟ 2.22 ⎜ P ⎟ Q = H3/ 2 ⎝ w ⎠ (1+ H)1/ 2 123 ExampleExample 10.1010.10 SolutionSolution3/33/3 124 UnderflowUnderflow GatesGates 1/41/4 A variety of underflow gate structure is available for flowrate control at the crest of an overflow spillway, or at the entrance of an irrigation canal or river from a lake. Three variations of underflow gates: (a) vertical gate, (b) radial gate, (c) drum gate. 125 UnderflowUnderflow GatesGates 2/42/4 The flow under the gate is said to be free outflow when the fluid issues as a jet of supercritical flow with a free surface open to the atmosphere. In such cases it is customary to write this flowrate as q = Cda 2gy1 (35) Where q is the flowrate per unit width. The discharge coefficient, Cd, is a function of contraction coefficient, Cc = y2/a, and the depth ration y1/a. 126 UnderflowUnderflow GatesGates 3/43/4 Typical discharge coefficients for underflow gates 127 UnderflowUnderflow GatesGates 4/44/4 The depth downstream of the gate is controlled by some downstream obstacle and the jet of water issuing from under the gate is overlaid by a mass of water that is quite turbulent. Drowned outflow from a sluice gate. 128 ExampleExample 10.1110.11 SluiceSluice gategate Water flows under the sluice gate shown in Fig. E10.11. The channel width is b = 20 ft, the upstream depth is y1= 6 ft, and the gate is a = 1.0 ft off the channel bottom. Plot a graph of flowrate, Q, as a function of y3. 129 ExampleExample 10.1110.11 SolutionSolution1/21/2 (35) q = bq = baCd 2gy1 = 393Cd cfs (Figure 10.29) Along the vertical line y1/a=6. For y3=6 ft, Cd=0 The value of Cd increases as y3/a decreases, reading a maximum of Cd=0.56 when y3/a=3.2. Thus with y3=3.2a=3.2ft Q = 393(0.56) cfs = 220 cfs For y3 < 3.2 ft the flowrate is independent of y3, and the outflow is a free outflow. 130 ExampleExample 10.1110.11 SolutionSolution2/22/2 The flowrate for 3.2ft ≤ y3 ≤ 6ft 131