1. Reservoirs, Spillways, & Energy Dissipators
CE154 – Hydraulic Design
Lecture 3
Fall 2009
CE154

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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3. Hoover Dam – downstream face
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4. Hoover Dam – Lake Mead
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5. Hoover Dam – Spillway Crest
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6. Hoover dam – Outflow Channel
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7. Hoover Dam – Outlet Tunnel
60 ft diameter outlet tunnel
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8. Hoover Dam – Spillway
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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. 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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14. Reservoir Capacity Curve
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15. Spillway Discharge Rating
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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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17. Spillway Capacity vs. Surcharge
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18. Spillway Cost Analysis
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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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22. Sabo Dam, Japan – Drop Chute
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23. New Cronton Dam NY – Stepped Chute Spillway
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24. Sippel Weir, Australia – Drop Spillway
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25. Four Mile Dam, Australia – Ogee Spillway
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26. Upper South Dam, Australia – Ogee Spillway
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27. Winnipeg Floodway - Ogee
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28. Hoover Dam – Gated Side Channel Spillway
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29. Valentine Mill Dam - Labyrinth
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30. Ute Dam – Labyrinth Spillway
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31. Matthews Canyon Dam - Chute
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32. Itaipu Dam, Uruguay – Chute Spillway
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33. Itaipu Dam – flip bucket
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34. Pleasant Hill Lake – Drop Inlet (Morning Glory) Spillway
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35. Monticello Dam – Morning Glory
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36. Monticello Dam – Outlet - bikers heaven
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37. Grand Coulee Dam, Washington – Outlet pipe gate valve chamber
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38. Control structure – Radial Gate
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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 happens
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41. Overflow Spillway
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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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43. Overflow Spillway
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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
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45. 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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46. Overflow Spillway
Why is C increasing with P? - P=0, broad crested weir, C= 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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47. C vs. P/Ho
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48. C vs. He/Ho
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49. C. vs. 
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50. Downstream Apron Effect on C
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51. Tailwater Effect on C
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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 = 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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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 jump
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56. Hydraulic Jump
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57. Hydraulic Jump V2 y2 y1 V1 Lj
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58. 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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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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60. Type IV Stilling Basin – 2.5<Fr<4.5
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61. Stilling Basin – 2.5<Fr<4.5
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62. Stilling Basin – 2.5<Fr<4.5
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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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65. Type III Stilling Basin – Fr>4.5
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66. Type III Stilling Basin – Fr>4.5
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67. 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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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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69. Type II Stilling Basin – Fr>4.5
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70. Type II Stilling Basin – Fr>4.5
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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
Use Manning’s equation to determine the normal flow condition in the upstream channel. V = 1.486R2/3S1/2/n Q = R2/3S1/2A/n A = 20y R = A/P = 20y/(2y+20) = 10y/(y+10) Q = = 1.486(10y/(y+10))2/3S1/220y/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= R2/3S1/2/n = V=Q/A = 400/22.2 = 18.02
Fr1 = V/(gy)1/2 =  a type IV basin may be appropriate, but first let’s check the tailwater level
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74. Example
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
A simple basin with an end sill may work well.
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75. Example
Length of basin Use chart on Slide #62, for Fr1 = 3.0, L/y2 = 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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