Transcript Reservoirs, Spillways, & Energy Dissipators

Reservoirs, Spillways, & Energy
Dissipators
CE154 – Hydraulic Design
Lecture 3
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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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Hoover Dam – downstream face
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Hoover Dam – Lake Mead
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Hoover Dam – Spillway Crest
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Hoover dam – Outflow Channel
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Hoover Dam – Outlet Tunnel
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Hoover Dam – Spillway
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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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Necessary Data
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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
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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
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Spillway Discharge Rating
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Spillway Design Procedure
• Route 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 process
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Spillway Capacity vs. Surcharge
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Spillway Cost Analysis
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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
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New Cronton Dam NY – Stepped Chute
Spillway
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Sippel Weir, Australia – Drop Spillway
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Four Mile Dam, Australia – Ogee
Spillway
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Upper South Dam, Australia – Ogee
Spillway
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Winnipeg Floodway - Ogee
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Hoover Dam – Gated Side Channel
Spillway
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Valentine Mill Dam - Labyrinth
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Ute Dam – Labyrinth Spillway
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Matthews Canyon Dam - Chute
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Itaipu Dam, Uruguay – Chute Spillway
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Itaipu Dam – flip bucket
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Pleasant Hill Lake – Drop Inlet (Morning
Glory) Spillway
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Monticello Dam – Morning Glory
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Monticello Dam – Outlet - bikers heaven
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Grand Coulee Dam, Washington – Outlet
pipe gate valve chamber
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Control structure – Radial Gate
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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
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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
Q  CL H e
3/ 2
C  f ( P, H e
H
,  , downstreamsubm ergence)
o
L  effectivewidth of spillway
H
H
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 total energyhead over crest
o
 designenergyhead overcrest
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Overflow Spillway
• Effective 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. 368-369
L  L  2( N K p  K a ) H e
'
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Overflow Spillway
• Discharge coefficient C
C = 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
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C. vs. 
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Downstream Apron Effect on C
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Tailwater Effect on C
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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 9-27
hd+d = 1.7×16 = 27.2’
P/Ho = 5/16 = 0.31
Co = 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 ft
hd = 20.7 ft
• Allowing 10% reduction in Co, hd+d/He =
1.2
hd+d = 1.2×16 = 19.2
Saving in excavation = 27.2 – 19.2 = 8 ft
Economic 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 flow
Fr < 1 – subcritical flow
• Transition from supercritical to
subcritical on a mild slope – hydraulic
jump
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Hydraulic Jump
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Hydraulic Jump
V2
y1
y2
V1
Lj
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Hydraulic Jump
• Jump in horizontal rectangular channel
y2/y1 = ½ ((1+8Fr12)1/2 -1) - see figure
y1/y2 = ½ ((1+8Fr22)1/2 -1)
• Loss of energy
E = E1 – E2 = (y2 – y1)3 / (4y1y2)
• Length of jump
Lj  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
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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/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/n
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Example
• Solve 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 = 18.02
• 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+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
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 sill
Use 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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