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Reservoirs, Spillways, & Energy Dissipators
CE154 – Hydraulic DesignLecture 3
Fall 2009 1CE154
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
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
CE154
Fall 2009 10
Necessary Data
• Location and site map• Hydrologic data• Climatic data• Geological data• Water demand data• Dam site data (foundation, material,
tailwater)
CE154
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
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
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
Fall 2009 13CE154
Reservoir Capacity Curve
Fall 2009 14CE154
Spillway Discharge Rating
Fall 2009 15CE154
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
Fall 2009 16CE154
Spillway Capacity vs. Surcharge
Fall 2009 17CE154
Spillway Cost Analysis
Fall 2009 18CE154
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
Fall 2009 19CE154
Spillway Design Procedure (cont’d)
• Perform hydraulic design of spillway structures- Control structure
- Discharge channel
- Terminal structure
- Entrance and outlet channels
Fall 2009 20CE154
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
Fall 2009 21CE154
Sabo Dam, Japan – Drop Chute
Fall 2009 22CE154
New Cronton Dam NY – Stepped Chute Spillway
Fall 2009 23CE154
Sippel Weir, Australia – Drop Spillway
Fall 2009 24CE154
Four Mile Dam, Australia – Ogee Spillway
Fall 2009 25CE154
Upper South Dam, Australia – Ogee Spillway
Fall 2009 26CE154
Winnipeg Floodway - Ogee
Fall 2009 27CE154
Hoover Dam – Gated Side Channel Spillway
Fall 2009 28CE154
Valentine Mill Dam - Labyrinth
Fall 2009 29CE154
Ute Dam – Labyrinth Spillway
Fall 2009 30CE154
Matthews Canyon Dam - Chute
Fall 2009 31CE154
Itaipu Dam, Uruguay – Chute Spillway
Fall 2009 32CE154
Itaipu Dam – flip bucket
Fall 2009 33CE154
Pleasant Hill Lake – Drop Inlet (Morning Glory) Spillway
Fall 2009 34CE154
Monticello Dam – Morning Glory
Fall 2009 35CE154
Monticello Dam – Outlet - bikers heaven
Fall 2009 36CE154
Grand Coulee Dam, Washington – Outlet pipe gate valve chamber
Fall 2009 37CE154
Control structure – Radial Gate
Fall 2009 38CE154
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
Fall 2009 39CE154
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
Overflow Spillway
Fall 2009 41CE154
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
Fall 2009 42CE154
Overflow Spillway
overcrestheadenergydesign
crestoverheadenergytotal
spillwayofwidtheffectiveL
esubmergencdownstreamPfC
CLQ
HH
HHH
o
e
o
e
e
),,,(
2/3
Fall 2009 43CE154
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
'
Fall 2009 44CE154
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%
Fall 2009 45CE154
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
Fall 2009 46CE154
C vs. P/Ho
Fall 2009 47CE154
C vs. He/Ho
Fall 2009 48CE154
C. vs.
Fall 2009 49CE154
Downstream Apron Effect on C
Fall 2009 50CE154
Tailwater Effect on C
Fall 2009 51CE154
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
Fall 2009 52CE154
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!
Fall 2009 53CE154
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
Fall 2009 54CE154
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
Fall 2009 55CE154
Hydraulic Jump
Fall 2009 56CE154
Hydraulic Jump
yy11 VV11
VV22 yy22
LLjj
Fall 2009 57CE154
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
Fall 2009 58CE154
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
Fall 2009 59CE154
Type IV Stilling Basin – 2.5<Fr<4.5
Fall 2009 60CE154
Stilling Basin – 2.5<Fr<4.5
Fall 2009 61CE154
Stilling Basin – 2.5<Fr<4.5
Fall 2009 62CE154
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
Fall 2009 63CE154
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
Fall 2009 64CE154
Type III Stilling Basin – Fr>4.5
Fall 2009 65CE154
Type III Stilling Basin – Fr>4.5
Fall 2009 66CE154
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.
Fall 2009 67CE154
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
Fall 2009 68CE154
Type II Stilling Basin – Fr>4.5
Fall 2009 69CE154
Type II Stilling Basin – Fr>4.5
Fall 2009 70CE154
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).
Fall 2009 CE154 71
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
Fall 2009 CE154 72
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
Fall 2009 CE154 73
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.
Fall 2009 CE154 74
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
Fall 2009 CE154 75
