Download - Reservoirs, Spillways, & Energy Dissipators CE154 – Hydraulic Design Lecture 3 Fall 20091CE154
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Reservoirs, Spillways, & Energy Dissipators
CE154 – Hydraulic DesignLecture 3
Fall 2009 1CE154
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Fall 2009 2
Lecture 3 – Reservoir, Spillway, Etc.
• Purposes of a Dam- Irrigation- Flood control- Water supply- Hydropower- Navigation- Recreation
• Pertinent structures – dam, spillway, intake, outlet, powerhouse
CE154
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Fall 2009 3
Hoover Dam – downstream face
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Fall 2009 4
Hoover Dam – Lake Mead
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Fall 2009 5
Hoover Dam – Spillway Crest
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Fall 2009 6
Hoover dam – Outflow Channel
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Fall 2009 7
Hoover Dam – Outlet Tunnel
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Fall 2009 8
Hoover Dam – Spillway
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Fall 2009 9
Dam Building Project
• Planning- Reconnaissance Study- Feasibility Study- Environmental Document (CEQA in California)
• Design- Preliminary (Conceptual) Design- Detailed Design- Construction Documents (plans & specifications)
• Construction • Startup and testing• Operation
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Fall 2009 10
Necessary Data
• Location and site map• Hydrologic data• Climatic data• Geological data• Water demand data• Dam site data (foundation, material,
tailwater)
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Dam Components
• Dam - dam structure and embankment
• Outlet structure- inlet tower or inlet structure, tunnels, channels and outlet structure
• Spillway- service spillway- auxiliary spillway- emergency spillway
Fall 2009 11CE154
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Spillway Design Data
• Inflow 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 12CE154
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Spillway Design Data (cont’d)
• Reservoir storage curve - storage volume vs. elevation- developed from topographic maps- requires reservoir operation rules for modeling
• Spillway discharge rating curve
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Reservoir Capacity Curve
Fall 2009 14CE154
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Spillway Discharge Rating
Fall 2009 15CE154
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Spillway Design Procedure
• Route the flood through the reservoir to determine the required spillway sizeS = (Qi – Qo) t Qi determined from IDF hydrograph Qo determined from outflow rating curve S determined from storage rating curve- trial and error process
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Spillway Capacity vs. Surcharge
Fall 2009 17CE154
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Spillway Cost Analysis
Fall 2009 18CE154
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Spillway Design Procedure (cont’d)
• 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 discharge
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Spillway Design Procedure (cont’d)
• Perform hydraulic design of spillway structures- Control structure
- Discharge channel
- Terminal structure
- Entrance and outlet channels
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Types of Spillway
• Overflow type – integral part of the dam-Straight drop spillway, H<25’, vibration-Ogee spillway, low height
• Channel type – isolated from the dam-Side channel spillway, for long crest -Chute spillway – earth or rock fill dam- Drop inlet or morning glory spillway-Culvert spillway
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Sabo Dam, Japan – Drop Chute
Fall 2009 22CE154
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New Cronton Dam NY – Stepped Chute Spillway
Fall 2009 23CE154
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Sippel Weir, Australia – Drop Spillway
Fall 2009 24CE154
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Four Mile Dam, Australia – Ogee Spillway
Fall 2009 25CE154
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Upper South Dam, Australia – Ogee Spillway
Fall 2009 26CE154
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Winnipeg Floodway - Ogee
Fall 2009 27CE154
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Hoover Dam – Gated Side Channel Spillway
Fall 2009 28CE154
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Valentine Mill Dam - Labyrinth
Fall 2009 29CE154
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Ute Dam – Labyrinth Spillway
Fall 2009 30CE154
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Matthews Canyon Dam - Chute
Fall 2009 31CE154
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Itaipu Dam, Uruguay – Chute Spillway
Fall 2009 32CE154
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Itaipu Dam – flip bucket
Fall 2009 33CE154
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Pleasant Hill Lake – Drop Inlet (Morning Glory) Spillway
Fall 2009 34CE154
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Monticello Dam – Morning Glory
Fall 2009 35CE154
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Monticello Dam – Outlet - bikers heaven
Fall 2009 36CE154
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Grand Coulee Dam, Washington – Outlet pipe gate valve chamber
Fall 2009 37CE154
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Control structure – Radial Gate
Fall 2009 38CE154
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Free Overfall Spillway
• Control - Sharp crested - Broad crested- many other shapes and forms
• Caution - Adequate ventilation under the nappe- Inadequate ventilation – vacuum – nappe drawdown – rapture – oscillation – erratic discharge
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Overflow Spillway
• Uncontrolled 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 happens
Fall 2009 40CE154
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Overflow Spillway
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Overflow Spillway Geometry
• Upstream 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
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Overflow Spillway
overcrestheadenergydesign
crestoverheadenergytotal
spillwayofwidtheffectiveL
esubmergencdownstreamPfC
CLQ
HH
HHH
o
e
o
e
e
),,,(
2/3
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Overflow Spillway
• Effective width of spillway defined below, where
L = effective width of crestL’ = net width of crestN = number of piersKp = pier contraction coefficient, p. 368Ka = abutment contraction coefficient, pp. 368-369
HKKL eapNL )(2
'
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Overflow Spillway
• Discharge coefficient CC = f( P, He/Ho, , downstream submergence)
• Why is C increasing with He/Ho?He>Ho pcrest<patmospheric C>Co
• Designing using Ho=0.75He will increase C by 4% and reduce crest length by 4%
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Overflow Spillway
• Why 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 maximum
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C vs. P/Ho
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C vs. He/Ho
Fall 2009 48CE154
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C. vs.
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Downstream Apron Effect on C
Fall 2009 50CE154
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Tailwater Effect on C
Fall 2009 51CE154
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Overflow Spillway Example
• Ho = 16’• P = 5’• Design an overflow spillway that’s
not impacted by downstream apron • To have no effect from the d/s apron,
(hd+d)/Ho = 1.7 from Figure 9-27hd+d = 1.7×16 = 27.2’P/Ho = 5/16 = 0.31Co = 3.69 from Figure 9-23
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Example (cont’d)
• q = 3.69×163/2 = 236 cfs/ft• hd = velocity head on the apron• hd+d = d+(236/d)2/2g = 27.2
d = 6.5 fthd = 20.7 ft
• Allowing 10% reduction in Co, hd+d/He = 1.2hd+d = 1.2×16 = 19.2Saving in excavation = 27.2 – 19.2 = 8 ftEconomic considerations for apron elevation!
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Energy Dissipators
• Hydraulic Jump type – induce a hydraulic jump at the end of spillway to dissipate energy
• Bureau of Reclamation did extensive experimental studies to determine structure size and arrangements – empirical charts and data as design basis
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Hydraulic Jump energy dissipator
• Froude number
Fr = V/(gy)1/2
• Fr > 1 – supercritical flowFr < 1 – subcritical flow
• Transition from supercritical to subcritical on a mild slope – hydraulic jump
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Hydraulic Jump
Fall 2009 56CE154
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Hydraulic Jump
yy11 VV11
VV22 yy22
LLjj
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Hydraulic Jump
• Jump in horizontal rectangular channely2/y1 = ½ ((1+8Fr1
2)1/2 -1) - see figure y1/y2 = ½ ((1+8Fr2
2)1/2 -1)
• Loss of energyE = E1 – E2 = (y2 – y1)3 / (4y1y2)
• Length of jumpLj 6y2
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Hydraulic Jump
• Design 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 basin
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Type IV Stilling Basin – 2.5<Fr<4.5
Fall 2009 60CE154
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Stilling Basin – 2.5<Fr<4.5
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Stilling Basin – 2.5<Fr<4.5
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Type IV Stilling Basin – 2.5<Fr<4.5
• Energy loss in this Froude number range is less than 50%
• To increase energy loss and shorten the basin length, an alternative design may be used to drop the basin level and increase tailwater depth
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Stilling Basin – Fr>4.5
• When Fr > 4.5, but V < 60 ft/sec, use Type III basin
• Type III – chute blocks, baffle blocks and end sill
• Reason for requiring V<60 fps – to avoid cavitation damage to the concrete surface and limit impact force to the blocks
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Type III Stilling Basin – Fr>4.5
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Type III Stilling Basin – Fr>4.5
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Type III Stilling Basin – Fr>4.5
• Calculate impact force on baffle blocks:
F = 2 A (d1 + hv1) where F = 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.
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Type II Stilling Basin – Fr>4.5
• When Fr > 4.5 and V > 60 ft/sec, use Type II stilling basin
• Because baffle blocks are not used, maintain a tailwater depth 5% higher than required as safety factor to stabilize the jump
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Type II Stilling Basin – Fr>4.5
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Type II Stilling Basin – Fr>4.5
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Example
• A 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).
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Example
1. Use Manning’s equation to determine the normal flow condition in the upstream channel.V = 1.486R2/3S1/2/nQ = 1.486 R2/3S1/2A/nA = 20yR = A/P = 20y/(2y+20) = 10y/(y+10)Q = 400
= 1.486(10y/(y+10))2/3S1/220y/n
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Example
• Solve the equation by trial and errory = 1.11 ftcheck A=22.2 ft2, P=22.2, R=1.0
1.486R2/3S1/2/n = 18.07V=Q/A = 400/22.2 = 18.02
• Fr1 = V/(gy)1/2 = 3.01 a type IV basin may be appropriate, but first let’s check the tailwater level
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Example
2. For a simple hydraulic jump basin, y2/y1 = ½ ((1+8Fr1
2)1/2 -1) Now that y1=1.11, Fr1=3.01 y2 = 4.2 ftThis 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
3. A simple basin with an end sill may work well.
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Example• Length of basin
Use chart on Slide #62, for Fr1 = 3.0,L/y2 = 5.25 L = 42 ft.
• Height of end sillUse design on Slide #60,Height = 1.25Y1 = 1.4 ft
• Transition to the tailwater depth or optimization of basin depth needs to be worked out
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