Reservoirs, Spillways, & Energy DissipatorsCE154 Hydraulic DesignLecture 3Fall 2009*CE154

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Fall 2009*Lecture 3 Reservoir, Spillway, Etc.Purposes of a Dam - Irrigation - Flood control - Water supply - Hydropower - Navigation - RecreationPertinent structures dam, spillway, intake, outlet, powerhouseCE154

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Fall 2009*Hoover Dam downstream faceCE154

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Fall 2009*Hoover Dam Lake MeadCE154

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Fall 2009*Hoover Dam Spillway CrestCE154

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Fall 2009*Hoover dam Outflow ChannelCE154

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Fall 2009*Hoover Dam Outlet TunnelCE154

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Fall 2009*Hoover Dam Spillway CE154

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Fall 2009*Dam Building ProjectPlanning - Reconnaissance Study - Feasibility Study - Environmental Document (CEQA in California)Design - Preliminary (Conceptual) Design - Detailed Design - Construction Documents (plans & specifications)Construction Startup and testingOperationCE154

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Fall 2009*Necessary DataLocation and site mapHydrologic dataClimatic dataGeological dataWater demand dataDam site data (foundation, material, tailwater)CE154

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Dam ComponentsDam - dam structure and embankmentOutlet structure - inlet tower or inlet structure, tunnels, channels and outlet structureSpillway - service spillway - auxiliary spillway - emergency spillway

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Spillway Design DataInflow Design Flood (IDF) hydrograph - developed from probable maximum precipitation or storms of certain occurrence frequency - life loss use PMP - if failure is tolerated, engineering judgment cost-benefit analysis use certain return-period flood Fall 2009*CE154

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Spillway Design Data (contd)Reservoir storage curve - storage volume vs. elevation - developed from topographic maps - requires reservoir operation rules for modelingSpillway discharge rating curveFall 2009*CE154

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Reservoir Capacity CurveFall 2009*CE154

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Spillway Discharge RatingFall 2009*CE154

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Spillway Design ProcedureRoute the flood through the reservoir to determine the required spillway size S = (Qi Qo) t Qi determined from IDF hydrograph Qo determined from outflow rating curve S determined from storage rating curve - trial and error processFall 2009*CE154

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Spillway Capacity vs. SurchargeFall 2009*CE154

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Spillway Cost AnalysisFall 2009*CE154

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Spillway Design Procedure (contd)Select spillway type and control structure - service, auxiliary and emergency spillways to operate at increasingly higher reservoir levels - whether to include control structure or equipment a question of regulated or unregulated dischargeFall 2009*CE154

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Spillway Design Procedure (contd)Perform hydraulic design of spillway structures - Control structure - Discharge channel - Terminal structure - Entrance and outlet channelsFall 2009*CE154

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Types of SpillwayOverflow type integral part of the dam -Straight drop spillway, H

Sabo Dam, Japan Drop ChuteFall 2009*CE154

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New Cronton Dam NY Stepped Chute SpillwayFall 2009*CE154

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Sippel Weir, Australia Drop SpillwayFall 2009*CE154

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Four Mile Dam, Australia Ogee SpillwayFall 2009*CE154

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Upper South Dam, Australia Ogee SpillwayFall 2009*CE154

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Winnipeg Floodway - OgeeFall 2009*CE154

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Hoover Dam Gated Side Channel SpillwayFall 2009*CE154

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Valentine Mill Dam - LabyrinthFall 2009*CE154

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Ute Dam Labyrinth SpillwayFall 2009*CE154

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Matthews Canyon Dam - ChuteFall 2009*CE154

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Itaipu Dam, Uruguay Chute SpillwayFall 2009*CE154

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Itaipu Dam flip bucketFall 2009*CE154

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Pleasant Hill Lake Drop Inlet (Morning Glory) SpillwayFall 2009*CE154

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Monticello Dam Morning GloryFall 2009*CE154

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Monticello Dam Outlet - bikers heavenFall 2009*CE154

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Grand Coulee Dam, Washington Outlet pipe gate valve chamberFall 2009*CE154

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Control structure Radial GateFall 2009*CE154

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Free Overfall SpillwayControl - Sharp crested - Broad crested - many other shapes and formsCaution - Adequate ventilation under the nappe - Inadequate ventilation vacuum nappe drawdown rapture oscillation erratic dischargeFall 2009*CE154

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Overflow SpillwayUncontrolled Ogee Crest - Shaped to follow the lower nappe of a horizontal jet issuing from a sharp crested weir - At design head, the pressure remains atmospheric on the ogee crest - At lower head, pressure on the crest is positive, causing backwater effect to reduce the discharge - At higher head, the opposite happensFall 2009*CE154

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Overflow SpillwayFall 2009*CE154

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Overflow Spillway GeometryUpstream Crest earlier practice used 2 circular curves that produced a discontinuity at the sharp crested weir to cause flow separation, rapid development of boundary layer, more air entrainment, and higher side walls - new design see US Corps of Engineers Hydraulic Design Criteria III-2/1 Fall 2009*CE154

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Overflow SpillwayFall 2009*CE154

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Overflow SpillwayEffective width of spillway defined below, where L = effective width of crest L = net width of crest N = number of piers Kp = pier contraction coefficient, p. 368 Ka = abutment contraction coefficient, pp. 368-369Fall 2009*CE154

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Overflow SpillwayDischarge coefficient C C = f( P, He/Ho, , downstream submergence) Why is C increasing with He/Ho? He>Ho pcrestCoDesigning using Ho=0.75He will increase C by 4% and reduce crest length by 4%Fall 2009*CE154

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Overflow SpillwayWhy is C increasing with P? - P=0, broad crested weir, C=3.087 - P increasing, approach flow velocity decreases, and flow starts to contract toward the crest, C increasing - P increasing still, C attains asymptotically a maximumFall 2009*CE154

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C vs. P/HoFall 2009*CE154

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C vs. He/HoFall 2009*CE154

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C. vs. Fall 2009*CE154

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Downstream Apron Effect on CFall 2009*CE154

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Tailwater Effect on CFall 2009*CE154

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Overflow Spillway ExampleHo = 16P = 5Design an overflow spillway thats not impacted by downstream apron To have no effect from the d/s apron, (hd+d)/Ho = 1.7 from Figure 9-27 hd+d = 1.716 = 27.2 P/Ho = 5/16 = 0.31 Co = 3.69 from Figure 9-23Fall 2009*CE154

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Example (contd)q = 3.69163/2 = 236 cfs/fthd = velocity head on the apronhd+d = d+(236/d)2/2g = 27.2 d = 6.5 ft hd = 20.7 ftAllowing 10% reduction in Co, hd+d/He = 1.2 hd+d = 1.216 = 19.2 Saving in excavation = 27.2 19.2 = 8 ft Economic considerations for apron elevation!Fall 2009*CE154

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Energy DissipatorsHydraulic Jump type induce a hydraulic jump at the end of spillway to dissipate energyBureau of Reclamation did extensive experimental studies to determine structure size and arrangements empirical charts and data as design basisFall 2009*CE154

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Hydraulic Jump energy dissipatorFroude number Fr = V/(gy)1/2 Fr > 1 supercritical flow Fr < 1 subcritical flow Transition from supercritical to subcritical on a mild slope hydraulic jumpFall 2009*CE154

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Hydraulic Jump Fall 2009*CE154

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Hydraulic Jumpy1V1V2y2LjFall 2009*CE154

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Hydraulic JumpJump in horizontal rectangular channel y2/y1 = ((1+8Fr12)1/2 -1) - see figure y1/y2 = ((1+8Fr22)1/2 -1)Loss of energy E = E1 E2 = (y2 y1)3 / (4y1y2)Length of jump Lj 6y2 Fall 2009*CE154

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Hydraulic JumpDesign guidelines - Provide a basin to contain the jump - Stabilize the jump in the basin: tailwater control - Minimize the length of the basin to increase performance of the basin - Add chute blocks, baffle piers and end sills to increase energy loss Bureau of Reclamation types of stilling basinFall 2009*CE154

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Type IV Stilling Basin 2.5

Stilling Basin 2.5

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Type IV Stilling Basin 2.5

Stilling Basin Fr>4.5When Fr > 4.5, but V < 60 ft/sec, use Type III basinType III chute blocks, baffle blocks and end sillReason for requiring V

Type III Stilling Basin Fr>4.5Fall 2009*CE154

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Type III Stilling Basin Fr>4.5Fall 2009*CE154

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Type III Stilling Basin Fr>4.5Calculate impact force on baffle blocks: F = 2 A (d1 + hv1) whereF = force in lbs = unit weight of water in lb/ft3 A = area of upstream face of blocks in ft2 (d1+hv1) = specific energy of flow entering the basin in ft. Fall 2009*CE154

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Type II Stilling Basin Fr>4.5When Fr > 4.5 and V > 60 ft/sec, use Type II stilling basinBecause baffle blocks are not used, maintain a tailwater depth 5% higher than required as safety factor to stabilize the jumpFall 2009*CE154

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Type II Stilling Basin Fr>4.5Fall 2009*CE154

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Type II Stilling Basin Fr>4.5Fall 2009*CE154

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ExampleA rectangular concrete channel 20 ft wide, on a 2.5% slope, is discharging 400 cfs into a stilling basin. The basin, also 20 ft wide, has a water depth of 8 ft determined from the downstream channel condition. Design the stilling basin (determine width and type of structure). Fall 2009CE154*

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ExampleUse Mannings equation to determine the normal flow condition in the upstream channel. V = 1.486R2/3S1/2/n Q = 1.486 R2/3S1/2A/n A = 20y R = A/P = 20y/(2y+20) = 10y/(y+10) Q= 400 = 1.486(10y/(y+10))2/3S1/220y/nFall 2009CE154*

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ExampleSolve the equation by trial and error y = 1.11 ft check A=22.2 ft2, P=22.2, R=1.0 1.486R2/3S1/2/n = 18.07 V=Q/A = 400/22.2 = 18.02Fr1 = V/(gy)1/2 = 3.01 a type IV basin may be appropriate, but first lets check the tailwater levelFall 2009CE154*

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ExampleFor a simple hydraulic jump basin, y2/y1 = ((1+8Fr12)1/2 -1) Now that y1=1.11, Fr1=3.01 y2 = 4.2 ft This is the required water depth to cause the jump to occur. We have a depth of 8 ft now, much higher than the required depth. This will push the jump to the upstream A simple basin with an end sill may work well. Fall 2009CE154*

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ExampleLength of basin Use chart on Slide #62, for Fr1 = 3.0, L/y2 = 5.25 L = 42 ft.Height of end sill Use design on Slide #60, Height = 1.25Y1 = 1.4 ftTransition to the tailwater depth or optimization of basin depth needs to be worked out Fall 2009CE154*

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******60 ft diameter outlet tunnel**************************************************************