This document provides an overview of reservoirs, spillways, and energy dissipators for dams. It discusses the purposes of dams and pertinent structures like dams, spillways, intakes and outlets. It describes the planning, design, construction, and operation process for dam projects. Key considerations for spillway design include inflow hydrographs, reservoir storage curves, discharge rating curves, and routing floods through the reservoir. Common types of spillways are discussed along with design procedures. Energy dissipation methods like hydraulic jump basins are also covered, including different basin types and design guidelines. An example problem demonstrates spillway and stilling basin design.
9. 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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10. 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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11. 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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12. 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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13. 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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16. 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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19. 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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20. 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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21. 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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39. 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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40. 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 happensFall 2009 40CE154
42. 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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44. 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
HKKL eap
NL )(2
'
+−=
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45. Overflow Spillway
• Discharge coefficient C
C = f( P, He/Ho, θ, 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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46. 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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52. 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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53. 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 apronFall 2009 53CE154
54. 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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55. 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
jumpFall 2009 55CE154
58. Hydraulic Jump
• Jump in horizontal rectangular channel
y2/y1 = ½ ((1+8Fr1
2
)1/2
-1) - see figure
y1/y2 = ½ ((1+8Fr2
2
)1/2
-1)
• Loss of energy
∆E = E1 – E2 = (y2 – y1)3
/ (4y1y2)
• Length of jump
Lj ≅ 6y2
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59. 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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63. 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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64. 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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67. Type III Stilling Basin – Fr>4.5
• Calculate impact force on baffle blocks:
F = 2 γ A (d1 + hv1)
where F = force in lbs
γ = 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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68. 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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71. 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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72. Example
1. Use Manning’s equation to determine
the normal flow condition in the
upstream channel.
V = 1.486R2/3
S1/2
/n
Q = 1.486 R2/3
S1/2
A/n
A = 20y
R = A/P = 20y/(2y+20) = 10y/(y+10)
Q= 400
= 1.486(10y/(y+10))2/3
S1/2
20y/n
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73. 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
Fall 2009 CE154 73
74. 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 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.Fall 2009 CE154 74
75. 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 outFall 2009 CE154 75