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

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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

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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

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

Spillway Discharge Rating

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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

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

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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

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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!

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).

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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

Fall 2009 CE154 75